This chapter is intended as a quick guide to the basic techniques of network analysis. The mathematical requirements are minimal; you need to be able to transpose equations and solve simultaneous equations by substitution. We also get to consider some characteristics of real voltage and current sources, which ties in nicely with some of the analysis tools presented.

We will take DC circuits for all of our examples, as this is the easiest way of demonstrating the techniques. Everything said applies equally to AC circuits as well, but you generally need to use complex numbers (see Chapter 4 and Appendix 4) to take account of phase if there are capacitors and inductors present, or sources at different phases.

There are two basic shapes that electric and electronic circuits come in; series and parallel. Each has associated with it one of two electrical laws which tell us how to go about calculating the voltages and currents in that circuit.

A series circuit is represented in Figure 5.1. The components are an input voltage, *V*,10 V DC, and three resistors: *R*_{1}, 1k; *R*_{2}, 2k2; and *R*_{3}*,* 4k7. We have also labelled on the drawing, for our convenience, the voltages across each resistor, *V*_{1,} *V*_{2} and *V*_{3}. The voltages are labelled with arrows. With DC voltages we follow a convention of putting the head of the arrow to the more positive end of the component.

The key to analysing a series circuit is to remember that the current is identical through each component in the loop. (We can also say that connecting the components in a different order will not affect that current.) Hence the current in the loop is labelled just once, this time with an arrow on the loop, with the legend *I* next to it to identify it. The head of the arrow shows the direction of current flow.

It is worth taking a moment to think about the rules that we apply when adding voltages and currents in using Kirchhoff’s two theorems, as it can be surprisingly easy to confuse ourselves.

Here’s the rules. If we follow these, things work:

1. The arrows for voltage across and current through a voltage or current source must point in the same direction.

2. The arrows for voltage across and current through a resistance must point in opposite directions.

3. For a voltage loop, the sum of arrows going clockwise equals the sum of arrows going anti-clockwise.

4. For a node, the sum of currents entering a node equals the sum of currents leaving the node, as denoted by the arrows’ directions.

In the above examples, it was obvious which end of the resistors would be more positive and in which direction currents would be flowing. Hence drawing the arrows was easy.

In other instances it will not be obvious, when labelling currents and voltages at the start of the analysis, which end of the component is the more positive. Such a situation is shown in Figure 5.3. We wish to know the voltage *V*_{2}*,* but this voltage is affected by both voltage and current sources which have a tendency to create voltages of opposite polarity across *R*_{2}. The way to approach this sort of problem is not to worry about the polarity of the voltage; if we label it in the wrong direction, the voltage will come out as negative in solving the equations. We simply get on with labelling the voltages and currents (following rules 1 and 2), then we can write equations for voltage loops (using rule 3, the voltage law) or nodes (with rule 4, the current law).

The voltage and current sources that we considered above were ‘ideal’ – that is, we consider that they will supply their correct voltage or current, regardless of what load is placed on them. (As such, they have no real existence.) For the ideal voltage source, we could say that the change in voltage across it is zero for any change in output current. This means that a notional quantity which we call its ‘internal resistance’ is also zero, by Ohm’s Law. For the current source we could say that the change in output current is zero for any change in output voltage. Hence its internal resistance is infinite. Real voltage and current sources are not ideal, and we have a way to represent this in our circuits (see Figure 5.5).

We give our ideal voltage source a series resistor, *R*_{s}. The closer this resistor is to a short circuit the closer the source is to ideal. If we connect a real voltage source to too low a load resistance the voltage across the load will drop (and the source could be damaged).

The current source is given a parallel resistor, *R*_{P}; the closer this is to an open circuit, the better the source. If we connect a real current source to too high a load resistance the current through the load will drop.

These circuit representations of real voltage and current sources are invaluable for circuit analysis, as we will now see. It is possible for any network of voltage sources, current sources and resistances to be replaced by either a single ideal voltage source with a series resistor, or a single ideal current source with a parallel resistor.

The techniques that we used in the last example of circuit analysis are fine for instances where we can get a solution for the circuit by writing two or three simultaneous equations and solving them. If the network gets much more complicated than that, however, the amount of maths involved can get pretty tedious, and it can get time consuming to arrive at the correct result without making a numerical error. Also, we have only solved the network for one value of load resistance; if we want to see what happens with a different value, we must do the calculations again. This is where replacing whole lumps of circuit with an equivalent non-ideal current or voltage source is great.

Here’s how we go about finding the right values for our equivalent circuit:

Replace with a voltage source (Thévenin’s equivalent circuit):

1. Cut the network at two places to isolate the section to be replaced.

2. Calculate the voltage at the load terminals with no load connected (called the ‘open circuit voltage’ or *V*_{TH}).

3. Calculate the equivalent resistance looking into the load terminals when any voltage sources are replaced with a short circuit and any current sources with an open circuit (called the ‘output resistance’ or *R*_{TH}).

The network can now be replaced with an ideal voltage source equivalent to *V*_{TH}, in series with a resistor equivalent to *R*_{TH.}

Replace with a current source (Norton’s equivalent circuit):

1. Cut the network at two places to isolate the section to be replaced.

2. Calculate the current at the load terminals when they are short circuited (called the ‘short circuit current’ or *I*_{N}).

3. Calculate the equivalent resistance looking into the load terminals when any voltage sources are replaced with a short circuit and any current sources with an open circuit (called the ‘output resistance’ or *R*_{N}).

The network can now be replaced with an ideal current equivalent to *I*_{N}, in parallel with a resistor equivalent to *R*_{N}.

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