8. Electrical networks

An electrical network can be considered as a graph in which a real number - called resistance - has been assigned to each edge .

In an electrical network if there is a potential difference between the endvertices of say a and b, then an electrical current - - will flow in the edge from a to b according to the Ohm's law:

An electrical network can be modelized by a directed multigraph where each edge is oriented arbitrary from one vertex to the other so, that we may use to denote the potential difference in the edge , representing the difference between the initial vertex and the endvertex. The notation denotes the current in the edge in the direction of the edge. A negative current will be considered as a positive current in the other direction.

In case of many practical problems electrical currents are constructed to enter the network at some points and leave it at others, and we interested in the consequent currents and potential differences in the edges.

To solve the problem we need use the Kirchoff's laws:

Kirchoff's potential (or voltage) law: Let be a cycle in G with vertices . Then in the potential differences round sum is 0:

Kirchoff's current law states that the current outflow from any vertex v is 0:

Here are the edges incident with v, and denotes the amount of current that leaves the network at v. In keeping with our convention, is the amount of current entering the network at v.

So, for vertices not connected to external points we have

If we know the resistances, then the potential law can be rewritten as a restriction on the currents in the edges. So, the currents are governed by the Kirchoff laws only; the physical characteristics of a network (the resistances) affect only the parameters in these laws. The potential law is equivalent to saying that one can assign absolute potentials to the vertices so that the potential difference between x and y is .

Consequence 1: If the network is connected and the potential differences are given for the edges, then we are free to choose arbitrarily the potential of one of these vertices, say , but then all the other potentials are determined.

As an example. we will consider the case In the most fundamental problems current is only allowed to enter the network at a single point s, called source, and only leave it at another point t, called sink. If the size of the current from s to t is w and the potential difference between s and t is p, then by Ohm's law is the total resistance of the network between s and t.

Let us consider the following example:

Figure 8.1. Network with 5 resistors

Our network has 5 resistors, of values 1,2,3,4 and 5 ohms. If we suppose that a unit current flow flows into the system at s and leaves it at t, then - for suitable values of e and f - the consequent edge currents must be as follows.

Figure 8.2. Currents in the example network

If we suppose that the potential at the sink is zero, i.e. , and we know that the potentials assigned to the vertices must satisfy Ohm's law , then we get

Therefore the potentials at the vertices are

  Figure 8.3. Potentials in the example network

On the other hand, Ohm's law has to be satisfied in more two edges, ab and bs. So we get

and

.

So we get

and

,

giving

and

.

In particular, the total resistance from s to t is

.

Since the Kirchoff's equations are linear and homogenous in all currents and potential differences, so we have two - important - consequences:

Consequence 2: The principle of superposition is valid for solutions.

Consequence 3: Any current resulting from multiple sources and sinks can be obtained by superposing flows belonging to one source and one sink.

Theorem 8.1.: The principle of superposition implies that in case of multiple sources and sinks there is at most one solution, no matter how the sources and sinks are distributed.

Proof.

Suppose we have two different solutions.

Because of Consequence 1, the difference of these solutions is also a solution, and this flow no current enters or leaves the network at any point.

Suppose that in this flow there is a positive current in some edge from a to b.

Then, by the current law, a positive current must go from b to c, then from c to d, etc., giving a trail abcd....

 

Since the network is finite, this trail has to return to a point previously visited.So we obtain a circuit in whose edges positive currents flow in one direction. It implies that the potential of each vertex is strictly greater than that of the next one round the circuit, and this is impossible.

How can we calculate the total resistance of a network? Unless the networks are very small, the calculations can get very heavy, and electrical engineers have a number of standard tricks to make them easier. Now, we will show some easy calculations for two basic-type networks:

  Figure 8.4. Resistors connected in series and in parallel.

Let us put a current of size 1 through the networks from s to t. We will calculate the total resistances in both cases.

In case of series connection:

and so, the total resistance is

In case of parallel connection we suppose that a current of size e goes through the first resistor and so, a current of size through the second. So

and so,

The total resistance is given by

If we define the conductance as the reciprocal of resistance then we can state the following theorem.

Theorem 8.2.: For series connection the resistances add and for parallel connection the con-ductances add.

In the following we will show how can we use convenient the conductances. Let us examine certain limiting cases of Ohm's law:

  • If the resistance of an edge ab is 0, then we necessarily have , and from an electrical point of view the vertices can be regarded as identical. (We say sometimes that a has been "shorted" to b.) Of course, a may be shorted to b if there is some other reason why .
  • If an edge has a conductance 0, then it does not have any effect to the currents and the potential. (We can produce such a situation if we "cut" an edge.)

Now, we will show an example how can we use a possible shorting of vertices to determine the total resistance of an electrical network. We take the network formed by the edges of a cube, in which each edge has 1 ohm resistance.

  Figure 8.5. A network formed by the edges of a cube

We would like to know what is the total resistance across an edge st?

Considering the Figure 8.5. we see that and , so a can be shorted to c and f can be shorted to d. So we get the following electrical network, where each resistance is equal to 1.

  Figure 8.6. Shorted edges in the example network

Now, for all the parallel resistors we can use the conductance addition rule, so we get the network at the next picture.

  Figure 8.7. The network after using the "conductance additivity".

From now on we can simplify resistors connected parallel and in series, until we find the total resistance is .

 

 

  Figure 8.8. Steps to get the total resistance