HUELLA CARBONO EN LA CADENA DE TRANSFORMACIÓN DE CACAO

There’s not too much to buy for this chapter of Starting Elec- tronics. As usual there’s a small number of resistors:

● 2 x lk5

● 1 x 4k7

● 2 x 100 k

As before, power ratings and tolerance of these resistors are not important.

We’re looking at current and voltage in our experiments in this chapter, so you’ll need to have a voltage and current source. The easiest and cheapest method is a simple battery. PP3, PP6, or PP9 sizes are best — as we need a 9 V source.

Whichever battery you buy, you’ll also need its correspond- ing battery connectors. These normally come as a pair of push-on connectors and coloured connecting leads. The red lead connects to the positive battery terminal; the black lead connects to the negative terminal.

The ends of the connecting leads furthest from the battery are normally stripped of insulation for about the last few mil- limetres or so, and tinned. Tinning is the process whereby the loose ends of the lead are soldered together. If the leads are tinned you’ll be able to push them straight into your breadboard. If the leads aren’t tinned, on the other hand, don’t just push them in because the individual strands of wire may break off and jam up the breadboard. Instead, tin them yourself using your soldering iron and some multicored solder. The following steps should be adhered to:

1) twist the strands of the leads between your thumb and first finger, so that they are tightly wound, with no loose strands,

2) switch on the soldering iron. When it has heated up, tin the iron’s tip with multicore solder until the end is bright and shiny with molten solder. If necessary, wipe off excess solder or dirt from the tip, on a piece of damp sponge; keep a small piece of sponge just for this purpose,

3) apply the tip of the iron to the lead end, and when the wires are hot enough apply the multicore solder. The solder should flow over the wires smoothly. Quickly remove the tip and the multicore solder, allowing the lead to cool natu- rally (don’t blow on it as the solder may crack if it cools too quickly). The lead end should be covered in solder, but no excess solder should be present. If a small blob of solder has

formed on the very end of the lead, preventing the lead from being pushed into your breadboard, just cut it off using your side cutters.

We’ll start our study, this chapter, with a brief recap of the ideas we covered in the last chapter. We saw then that re- sistors in series may be considered as a single equivalent resistor, whose resistance is found by adding together the resistance ofeach resistor. This is called the law of series resistors, which is given mathematically by:

Similarly, there is a law of parallel resistors, by which the sin- gle equivalent resistance of a number of resistors connected in parallel is given by:

Using these two laws many involved circuits may be broken down, step by step, into an equivalent circuit consisting of only one equivalent resistor. The circuit in Figure 2.9 last chapter was one such example and, if you remember, your homework was to calculate the current I from the battery. To do this you first have to find the single equivalent resist- ance of the whole network, then use Ohm’s law to calculate the current.

Figure 3.1 shows the first stage in tackling the problem, by dividing the network up into a number of smaller networks. Using the two expressions above, associated with resistors in series and parallel, we can start to calculate the equivalent resistances of each small network as follows:

Figure 3.1 A resistor network in which the total resistance can only be calculated by breaking it down into blocks

Network A

Network A consists of two resistors in parallel. The equivalent resistance, R_A, may therefore be calculated from the above expression for parallel resistors. However, we saw in the last chapter that the resistance of only two parallel resistors is given by the much simpler expression:

Network B

Three equal, parallel resistors form this network. Using the expression for parallel resistors we can calculate the equiva- lent resistance to be:

which gives:

Hint:

This is an interesting result, as it shows that the equivalent resistance of a number of equal, parallel resistors may be easily found by dividing the resistance of one resistor, by the total number of resistors. In this case we had three, 15 k resistors, so we could simply divide 15 k by three to obtain the equivalent resistance. If a network has two equal resistors in parallel, the equivalent resistance is one half the resistance of one resistor (that is, divide by 2). If four parallel resistors form a network, the equivalent resistance is a quarter (that is, divide by 4) the resistance of one resistance of one resistor. Hmm, very useful. Must remember that _— right?

Network C

Using the simple expression for two, unequal parallel resis- tors:

Network D

The overall resistance of two series resistors is found by adding their individual resistances. Resistance R_D is there- fore 10 k.

We can now redraw the whole network using their equivalent resistances, as in Figure 3.2, and further simplify the result- ant networks.

Figure 3.2 The same network, broken down into smaller networks using equivalent resistances (networks E, F and G)

Network E

This equivalent resistance is found by adding resistances R_A and R_B. It is therefore 12k5.

Network F

Same as network E. Resistance R_F = 16 k.

Network G

Two equal parallel resistors, each of 10 k. Resistance R_G is therefore 5 k. Figure 3.3 shows further simplification.

Figure 3.3 The network further simplified (network H) into two unequal parallel resistances

Network H

Two unequal, parallel resistors. Resistance R_H is therefore given by:

Figure 3.4 shows a further simplification.

Network I

Resistance R_I is found by adding resistances of resistors R₁, R_H and R_G. Resistance R_I is therefore 22k02.

Figure 3.5 shows the next stage of simplification with two unequal parallel resistors. The whole network’s equivalent resistance is therefore given by:

Now, from Ohm’s law we may calculate the current I, flowing from the battery, using the expression:

Did you get the same answer?

Figure 3.5 The final simplification, into two parallel resistances. From this the current can be calculated