ELECTRICAL NETWORKS
We examine an application of the ideas developed in the previous section. This material is presented solely for the stimulation and motivation of the interested reader, and is not required reading for other parts of the text.
An electrical network is a physical system consisting of various kinds of “network elements” such as voltage sources (e.g., batteries or generators), resistors and capacitors, to name but a few. These are generally connected together by means of wires or other conductors of electricity. Of course radios, television sets, and computers are examples of complex electrical networks. The behaviour of an electrical network is characterized in terms of the electrical current i through each network element and the voltage v across each element. The current i is a measure of the actual member of electrons flowing through the element, while the voltage v across the element is, roughly speaking, a measure of the force driving the electrons through the element. Current and voltage are measured in units called amperes and volts respectively, and by instruments called ammeters and voltmeters respectively.
Network elements are characterized by the relationship between the voltage across them and the current through them. For example, the simplest element is a “resistor”, for which the relationship takes the form v = Ri, where R is constant called the resistance of the resistor. This relationship is known as Ohm’s Law after its discoverer, George Simon Ohm (1787- 1845), a German high-school teacher and experimenter. For other types of elements, the relationship generally involves calculus, so we restrict our attention to network involving only resistors and batteries. Two such networks are shown in figure 4-6. In the figure, the R, indicate resistors, while each circle labeled “v” is a battery of voltage v volts. They are connected by wires represented by line segments in the figure. Of course, the battery drives electrons through the circuit. Such circuits, through simple, are often found as part of larger networks.
Electrical networks such as these may be analysed by regarding them as graphs. When this is done, the conventional representations are simplified by omitting the straight-line segments (“wires”) connecting the elements to produce equivalent graphical representations. Thus, for example, the network of figure 4-6(a) yields a graph with 4 vertices and 5 edges as shown in figure 4-7(a).
In addition to Ohm’s law the equations that describe the behaviour of an electrical network such as the ones in the figures are determined by intuitively plausible physical principles known as Kirchhoff’s Laws, after Gustav Robert Kirchoff (1824-1887), an eminent German physicist. We illustrate these principles using the network of figure 4-6(a).
The graphical representation in figure 4-7(a) has been redrawn in figure 4-8 with a simplified notation for the analysis in terms of graphs.
First, Kirchhoff’s current law states that the sum of the currents flowing into any vertex must equal the sum of the currents flowing out. To implement this, we arbitrarily assign directions (arrows) of current flow through each element, thereby making the network into a directed graph. We arbitrarily adopt the convention that currents flowing into a vertex are given negative signs (respectively, positive signs for those flowing out). This does not affect the system of equations as a whole.
Let iz denote the current flow through element z. The current equations for this network are
vertex 1: ia + ie = 0
vertex 2: -ia + ib + id = 0
vertex 3: -ib + ic = 0
vertex 4: - ic - id - ie = 0
We leave as an exercise for the reader the verification that the matrix form for these equations is Aõ = 0, where A is the 4 x 5 incidence matrix of the directed graph and õis the 5 x 1 matrix of currents in the order labelled. Notice that the last equation is the sum of the first three; we say the equations are not independent. In fact, the rank of the incidence matrix A is 3, which is one less than the number of vertices. This observation is pursued in the exercise.
The second physical principle is Kirchhoff’s voltage law, which states that the algebraic sum of the voltages around any loop must be zero. A loop (or circuit) is a path (i.e., a sequence of contiguous edges) that begins and ends at the same vertex, and doesn’t pass through any vertex more than once. For example, in figure 4-8, there are three loops, namely, ade, bcd, and abce.
To implement the voltage law, we adopt the convention that current flows from (+) to (-) across each element. If this coincides with the direction of the loop, the voltage across the given element is taken to be positive, and negative if otherwise. Let vz denote the voltage across the element z. Then the voltage equations are (with vi = v = the voltage supplied by the battery)
loop ade: va + vd - ve = 0
loop bcd: vb + vc - vd = 0
loop abce: va + vb + vc - ve = 0
The coefficient matrix B of this system of equations is called a circuit matrix of the directed graph. In general, Kirchhoff’s voltage equations for a network can be written BV = 0, where B is a circuit matrix for the graph of the network and V is the column vector of voltages (it has entries va, vb, vc, vd, ve in the example above). The definition of B and some of its properties are discussed in the exercises. In particular, its rank is seen to be equal to the number of edges less the number of vertices, plus one.
To conclude, we apply these ideas to an analysis of the circuit in figure 4-8, where the elements have values v = ve = 10 volts, and R1 = 200, R2 = 200, R3 = 400, R4 = 600 ohms. Ohm’s law for the resistors yields the equations:
R1: va = 200ia
R2: vb = 200ib
R3: vd = 400id
R4: vc = 600ic
where voltages are in volts and currents in amperes. We substitute into the first two voltage equations to get
200ia 400id – 10 = 0
200ib + 600ic - 400id = 0
Together with the first three current equations, namely,
ia + ie = 0
-ia + ib + id = 0
-ib + ic = 0
we now have five linear equations in ia, ib, ic, id, ie with augmented matrix.
200 0 0 400 0 10
0 200 600 -400 0 0
1 0 0 0 1 0
-1 1 0 1 0 0
0 -1 1 0 0 0
The reader should verify that this is row-equivalent to
1 0 0 0 0 3/140
0 1 0 0 0 1/140
0 0 1 0 0 1/140
0 0 0 1 0 1/70
0 0 0 0 1 -3/140
Thus, the unique solution is ia = 3/140, ib = 1/140, ic = 1/140, id = 1/70, and ie = -3/140 (amperes). In addition, for example, the voltage across resistor R4 is vc = 600ic = 600/140 » 4.29 volts.
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