# Topology and Elementary Electric Circuit Theory, I

This column will look at the hidden role of the Euler characteristic in circuit theory... Tony Phillips
Stony Brook University
Email Tony Phillips

### Introduction

As a Sophomore at MIT I took the course 6.10, Elementary Electric Circuit Theory (required at that time for all would-be physics majors) from David Huffman. That course was in fact my first introduction to topological concepts even though the word "topology," to my recollection, was never mentioned. It does occur in a text written for this course a few years before: Introductory Circuit Theory by Ernst Guillemin (Wiley, NewYork, 1953). In this column and my next I plan to go over some of this material from the point of view of a topologist, to try and share some of the excitement I felt at seeing these abstract concepts emerge from a completely practical context. This column will look at the hidden role of the Euler characteristic in circuit theory; the next will examine duality. The Wheatstone Bridge (Sir Charles Wheatstone, 1843) is an electric circuit designed to measure an unknown resistance $R_x$. A voltage source (at left) is connected to an array of resistors. The variable resistance $R_2$ is adjusted to make the voltage between $B$ and $D$ (measured by the voltmeter $V_G$) equal to zero. Then $R_2/R_1 = R_x/R_3$. Image available through Creative Commons.

### A restricted class of circuits

In mathematical terms, an elementary electric circuit is a connected, directed graph. The graph is made of conducting wire, and each segment of the graph bears a device which can be a voltage source, a current source, a resistor ($R$), an inductance ($L$), or a capacitor ($C$).
Most of the phenomena related to topology also occur in circuits with no inductances or capacitors; here we will work just with these.
If the oriented segment $\langle ab \rangle$ bears a voltage source $V$, then the voltage difference $\Delta V_{ab}$ between the ends of the segment is equal to to $V$; if it bears a current source $I$ then it will carry a current $I$ from node $\langle a\rangle$ to node $\langle b\rangle$. Otherwise $\Delta V_{ab}$ and the current $I_{ab}$ from node $\langle a\rangle$ to node $\langle b\rangle$ are related by Ohm's Law:

$\Delta V_{ab} = R\,I_{ab}$.

Note that $\displaystyle{\Delta V_{ba}= -\Delta V_{ab}}$ and $I_{ba} = -I_{ab}$.

The voltage differences and currents in a circuit must satisfy Kirchhoff's Laws

• If segments $\langle a_1a_2 \rangle, \langle a_2a_3 \rangle, \dots \langle a_na_1 \rangle$ form a loop, then $\Delta V_{a_1a_2} + \Delta V_{a_2a_3} + \cdots + \Delta V_{a_na_1} = 0$ (KVL, Kirchoff's Voltage Law).
• If $\langle a_1b\rangle, \langle a_2b\rangle, \dots \langle a_nb\rangle$ lists all the segments sharing the node $\langle b\rangle$, then $I_{a_1b} + I_{a_2b} +\cdots I_{a_nb} =0$ (KCL, Kirchoff's Current Law).

Remark: if a circuit has $N$ nodes $b_1,\dots, b_N$, the sum of the left-hand-sides of all $N$ equations coming from KCL must automatically be zero, since each current appears twice, as $I_{ab}$ and $I_{ba}$, with opposite signs. It follows that the sum of the first $N-1$ equations is equivalent to the last one.

### Solving a circuit

Solving a circuit means calculating the currents in all the segments.

Elementary loops (meshes). A mesh in a circuit is a loop that contains no smaller loops.    One way to organize the solution of the network is in terms of mesh currents, the clockwise (say) currents around the boundary of each mesh. The current in a segment will be the algebraic sum of the currents from the two meshes it belongs to (one if the segment is on the outside). This gives a collection of unknowns $i_1, \dots, i_M$ where $M$ is the number of meshes.
Applying KVL to each of the meshes gives a set of $M$ equations which can be solved for $i_1, \dots, i_M$, thus yielding the currents in all the segments and the solution of the circuit. An example, adapted from Wikipedia, should show how the procedure works. This circuit has voltage sources of 3 and 4 Volts and resistors of 10, 20 and 30 Ohms linked as in the diagram. Applying KVL to the top mesh, we add the voltage increments as we go around the boundary of the mesh, starting at the node on the left: $-10 i_a - 20 (i_a-i_b) + 3 =0$; similarly the bottom mesh yields $-3 -20 (i_b-i_a) -30 i_b -4 = 0$. Hence the system of linear equations: $$-30i_a+20i_b = -3$$ $$20i_a-50i_b=7,$$

which give $i_a=1/110\, \mbox{A},\,\, i_b=-15/110\, \mbox{A}$ and so $i_1=i_a=1/110\, \mbox{A},\,\, i_2 =i_b-i_a=-16/110\, \mbox{A},\,\, i_3=-i_b=15/110\, \mbox{A}$. The circuit is solved.
A similar calculation applied to the Wheatstone Bridge proves $R_2/R_1 = R_x/R_3$ when voltage $BD$ is zero.

### Where is the topology?

(The Euler Characteristic). The straightforward calculation we just did contains a mathematical subtlety, because it only works for a planar graph, and this fact is not obvious. Guillemin addresses this problem (p. 54); he explains in particular that the term "mesh" should not refer to "the contour" but rather to "the space surrounded by that contour." In topological terms, the meshes must be considered as 2-dimensional elements of the circuit. If a circuit is planar, we can embed it on the surface of a sphere, and its graph will define a convex polyhedron. There will be one new mesh, corresponding to the outside of the planar graph, but no new equation since if all the mesh equations are added, their sum must be zero: each segment is counted twice with opposite signs; the new equation is the sum of the old ones. For a circuit on the sphere, the numbers $M$ of meshes, $S$ of segments, and $N$ of nodes are related by the Euler characteristic: $M-S+N=2$. It follows that the number $M-1$ of independent mesh equations plus the number of independent node equations ($N-1$, as remarked above) is equal to the number $S$ of unknown segment currents, allowing for a unique solution. This point is nicely explained in a video featuring Peter Lax. A circuit with the topology of the "utility graph" K(3,3), embedded in the torus.

For a non-planar circuit, Kirchhoff's laws do not produce a unique solution. For example, a circuit with the K(3,3) topology cannot be embedded in the plane but can be embedded in the torus, as shown in the figure above. There will be 2 independent mesh equations corresponding to ADBE and DBFC, and 5 independent node equations. Since there are 9 segments, this leaves two undetermined segment currents. These correspond to two independent non-bounding loops, say ADCEA and AFBDA. In fact currents in those loops could be produced (Faraday effect) by varying magnetic fluxes linking the torus inside or outside. $\Phi_1(t)$, $\Phi_2(t)$, variable flux linking circuit loop AFBDA, ADCEA, respectively. By Faraday's Law, a variable flux will produce a current in a loop it links.

(Homological analysis). A fancier explanation can be given in terms of homology, which will be useful when we study duality. As explained above, we can consider a planar network as embedded in the sphere, so that its nodes (the 0-cells), segments (1-cells), and meshes (2-cells) form a cell complex $X$ with total space the sphere.
The $k$-dimensional chain complex $C_k(X)$ (here $k=0, 1, 2$) is the vector space of formal linear combinations of the $k$-cells. The boundary homomorphisms $\partial\!: C_1(X) \rightarrow C_0(X)$, which takes the oriented 1-cell $\vec{AB}$ to the 0-chain $B - A$, and $\partial\!: C_2(X) \rightarrow C_1(X)$, which takes the mesh with positively oriented boundary $ABC\dots KA$ to the 1-chain $\vec{AB} + \vec{BC} +\cdots +\vec{KA}$, satisfy $\partial\partial=0$ since each endpoint is counted twice with opposite signs. A $k$-chain $c$ with $\partial c = 0$ is called a $k$-cycle, and one satisfying $c=\partial b$ for some $(k+1)$-chain $b$ is called a $k$-boundary. Since $\partial\partial=0$ the 1-boundaries of $X$ are a subspace of the 1-cycles; the quotient vector space is the first homology vector space of $X$.
A basic result in homology theory is that for a cell complex with total space the sphere, the first homology must be zero: every 1-cycle must be a 1-boundary.    In our circuit, let $c_1,\dots c_S$ be the segments, taken as oriented 1-cells, and consider the 1-chain $I = i_1c_1 + i_2c_2 + \dots i_Sc_S$ where the coefficient of each segment is its current, with the sign adjusted to get the flow in the right direction. Then Kirchhoff's Current Law becomes the statement that $\partial I = 0$ (the coefficient of each node is the algebraic sum of the incident currents, hence 0). Since the topology of the circuit forces every 1-cycle to be a boundary, there must be a 2-chain $J$ with $\partial J = I$. Writing $J=a_1e_1 + a_2e_2 +\cdots a_Me_M$ where $e_1, \dots e_M$ are the oriented 2-cells corresponding to the meshes of $X$, the coefficient $a_p$ is the mesh current associated to the $p$th mesh. This is the homological justification of the mesh-current method used in our example. (In general, the 2-chain $J$ could be varied by an arbitrary 2-cycle $H$, since $\partial (J+H)=\partial J + \partial H = \partial J = I$. In practical terms, increasing all the mesh currents by the same quantity, necessary for $\partial H = 0$, makes no difference in the segment currents they define, i.e. in the solution of the circuit.)

In part II, we'll examine duality in circuits.

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