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Resistance and Fields 
After our brief excursion into atomic theory in Chapter One, we get straight down to the practicalities here in Chapter Two. It deals with the quality called resistance and the field. 
I hope you have fully grasped the contents of the first section. If not please read it again. Everything from now on builds on that basis, and to cover the distance we have to speed up a bit. If anything seems confusing come back to it a bit later. Usually things make sense on the second or third reading.
Look at Figure 1 . Here we again have our two charged objects but this time a rather thick wire of conducting metal has been shoved between the two connecting them. What happens? What happens is that the electrons finding a nice conductor available whiz down it propelled by the electric field to even out the charge on the two objects [NOTE 1] . As we said, this flow of electrons is called an electric current, and as the current flows, there will be a magnetic effect. How fast will the electrons be able to neutralize the difference in charge? Well, the rate of flow through the wire will depend on various factors:
The difference in potential (we'll call it potential difference from now on) between the two charged surfaces
The resistance of the wire
The total difference in charge in electron'sworth.
Any conductor will show this quality of resistance to flow of electrons (there are exceptions called superconductors but you won't have any on the boat for a few years). To explain it further, look at Figure 2 . Here you see a tube that is stuffed full of pingpong balls (tennis balls if you prefer). There is no room for any more. So if we push a ball into one end, out from the other will pop another ball, leaving the tube with the same number as before (that is  full). We have put one ball in and got one ball out at the other end, but it is NOT the same ball. In exactly this way, when electrons are forced into a conductor the same quantity will come out at the far end, but NOT the same ones. Each electron has to displace an electron from an atom, which moves on and displaces an electron from another atom, and so on. This process uses energy, and is called resistance. The energy is used up in making the electrons move, and in doing so the atoms are caused to vibrate at a greater rate, which is to say that they get hotter. The resistance limits the rate at which electrons can move, and good conductors obviously will have a low resistance, whereas insulators have a very high resistance. How fast will all this happen? Look at the tube of balls. The ball will pop out at one end at the same time as we force the ball in to the other end, however long the tube (in principle!). In a conductor the electrons do indeed pop out at one end almost simultaneously with them entering at the other  but the individual electrons move much more slowly, perhaps a couple of centimetres a minute. The general slow movement of electrons under the influence of a field is called electron drift, which gives the idea. The time delay between the electrons entering and leaving is short, because the disturbance caused travels along the wire at up to 300,000 kilometres a second  adequately fast, with a short wire! Resistance depends upon the material and also the dimensions of the material. For an internally consistent material, if a particular piece has a particular resistance, two of them connected endtoend will have twice as much. Look at Figure 3 . Here we have a squaresectioned piece of wire. If we add another equal piece on the end, the resistance of the whole length will be twice that of the first piece. This means that resistances in line add up. If however we put the two sidebyside, the result will be a piece having half the resistance. By the same argument, if we quadruple the crosssectional area of the piece of wire (equivalent to doubling the diameter), the resistance will be one quarter of that of the original piece. Therefore resistances connected sidebyside reduce. Now this is easy enough to follow, isn't it? It holds true for all materials, insulators as well, so we can make some statements about resistance:
We could make a variable resistance quite easily  I'm sure you can work out how. If we take a length of material and then make a moving contact that can slide along it, we can effectively vary the resistance. We will come back to this.
The flow of electrons is called an electric current. We could, again in principle (given remarkable eyesight!), sit and watch a piece of wire with a current flowing in it and count the number of electrons passing a particular point in a second. We could then use a unit of the electronsecond. Electrons are very small, however, and so what we actually do is use a much larger unit than a single electron  we use a practical unit called the ampere (usually known as amp) which is (approximately) 6,240,000,000,000,000,000 electrons per second. It is actually defined in terms of the magnetic effect caused by this quantity of flow. Having got this unit, we can now define a quantity of electricity  that is a measure of total charge, a quantity of electrons. We could use the number of electrons that pass a particular point in one second when a current of one amp is flowing  that is one amperesecond. This unit is called a coulomb , but we won't be using it much. Once again there is a larger unit, which is the amperehour  this obviously is the number of electrons that pass a point in one hour when a current of one amp is flowing. As there are 3,600 seconds in an hour, the amperehour is equal to 3,600 coulombs and it has a symbol which is Ah.
As I said, the current flowing is affected by the potential difference and the resistance. Let's now choose a unit for potential difference. I have said that energy is used up in forcing electrons through a resistance  that is, work is being done. The unit for work is called the joule. I won't give you its definition here, as this is not a Physics text book, but it is roughly equivalent to lifting a weight of one hundred grams to a height of one metre. We can, given this information, define a unit for potential difference like this: there is a potential difference of one unit between two points if one joule of work is done when one coulombsworth of electricity flows from one to the other. The unit is actually called a volt. We can now define a resistance as having a value of one (something) if a potential difference of one volt causes a current of one amp to flow. One what? The name chosen is one ohm. All these units (and most of the others) are named after pioneers in the field: Messrs Volta, Ampére, Coulomb, Joule and Ohm [NOTE 2] .
Now, one volt is the electrical potential difference that will cause a current of one amp to flow through a resistance of one ohm.
Have a quick look at Figure 4 . One ohm is defined here as the resistance that will allow a current of one amp to flow if a potential difference of one volt is applied to it, and one amp is the current that flows when one volt is applied to a resistance of one ohm. If that's not circular I don't know what is! But it all works fine, based on the ampere, joule and coulomb [NOTE 3] .
Let us look at these units again:
These symbols, A,V, and Ω are for particular quantities of current, potential difference and resistance  we could have 2A (two amps) or 12V (twelve volts) or 1000Ω (one thousand ohms). We will also want to come up with some more generalized symbols to represent these quantities in the abstract, that is current, voltage and resistance in general. The symbols chosen are I for current, V for potential difference and R for resistance. It can be a little confusing that V represents both (a quantity of potential difference in the abstract) and (a particular potential difference in units of volts) but I'm sorry, I didn't choose the symbols, I just use them! Because of the way we have defined them, these are all interdependent and can be written in a formula: for a particular current, resistance, and voltage:
number of volts/number of amps=number of ohms
or, in general,
V/I=R
This very simple formula can be easily transposed to give any of the quantities as the result, and in Figure 5 I show a way I recommend that you do this, and how to remember the formulae. Let's look at an example or two: If we have a voltage of 10 volts and a current of 2 amps, what is the resistance?
V / I = R
V = 10 : I = 2 : 10 / 2 = 5 ohms
If we have a voltage of 13.8 volts and a resistor of 40 ohms, what is the current? V / I = R but we want I so use I = V / R
V = 13.8 : R = 40 : 13.8 / 40 = 0.345 amps
Easy, isn't it? This relationship is loosely known as Ohm's Law. It isn't really quite what he said. He found that these relationships are true for a long straight wire, or at least, they are true, so long as it doesn't get hot. Wire when it gets hot increases markedly in resistance, so the 'Law' wouldn't quite work. A material for which Ohm's Law works is called ohmic and a material that doesn't exhibit this linear relationship is called nonohmic.
So far we have been getting our current from a static electrical potential difference. Fortunately there are various techniques for making a potential difference between two points other than rubbing amber with fur, the principal ones being chemical means (cells and batteries) and mechanical means. We will go into how they work later, but for now we need to understand something called electromotive force. In rubbing an amber rod with fur, some work has been done in moving electrons from one place to another. This has resulted in an unequal distribution of electrons and that has resulted in the creation of an electric field. Let's say that we have ended up with a potential difference of 1000V (one thousand volts)  not unlikely. Now what if we wanted to use a battery to pump some more electrons to the negatively charged surface? There is an electric field trying to make electrons travel the other way, towards positive. If we connected up a typical 9V battery it won't get very far trying to move electrons to a place that is at a potential of 1000V. No, we would need a source with a potential of more than 1000V. What we need is something that can move electrons against the field of a potential difference of 1000V, an electron motive force or electromotive force (thankfully abbreviated to e.m.f.) of more than 1000V. An e.m.f. is measured in terms of the potential difference (I'll call it p.d. from now on if you don't mind) that it can overcome. This is fairly clear, but in case there is doubt let's use an analogy: see Figure 6 and Figure 7 . You are pumping out the bilges. If it is a deepdraught vessel and there is a sump, you may have to pump the water up let's say five feet (1.5 metres or so!). You will have to supply a motive force to the water at least equal to the height it has to go to, five feet. You could call it watermotive force if it is any clearer, and if you supply less than five feet'sworth the water will not get up to the top. In the same way, e.m.f. is specified as the electrical pressure it can overcome. It is measured in volts, because p.d. is measured in volts, and it will push electrons against a p.d. less than the e.m.f. Have you got that? If not, read it again and try the bilge pump. There will be another example later. From now on we will distinguish when necessary between e.m.f. and p.d., but do remember they are both measured in volts. The symbol for e.m.f. is the letter E, and it can be directly substituted in the formula above: E = I / R with E in volts, I in amps and R in ohms. In practice we will rarely use this formula, since e.m.f is measured in volts we just use the volts figure in V = I / R.
Before we go further we will look more closely at resistance and e.m.f. I am sure you know what a battery is. It provides a source of electricity, a source of e.m.f. This means it can force a current of electrons against a p.d., as I have explained. Take a very simple example. In Figure 8 you will see a simple cell (torch type) connected to a piece of wire. What happens? The battery is a source of e.m.f. and will force electrons from its negative to its positive terminal. The conventional direction of flow is unfortunately from the positive terminal to the negative terminal, but never mind. We know the wire has resistance, so there will be a current flowing through a resistance. How much current will flow? It depends on the e.m.f. available, and the resistance of the wire. If we know the e.m.f., which is usually printed on the battery, and if we know the resistance of the wire, we could calculate the current. But I said that e.m.f. is the capacity to force a current against a p.d., doing work. Where is the p.d.?? The p.d. is developed by the wire, or rather by the resistance of the wire. There is work being done in forcing the current of electrons through the wire, this work develops a p.d., and the system will settle down with the p.d. developed by the wire equal to the e.m.f. of the battery. From this point of view, the two labels e.m.f. and p.d. might as well be the same. But look at Figure 9 . Here we have two batteries, each of nominally the same voltage, connected positive to positive and negative to negative by two wires. What happens now? Well, if the e.m.f.s of the two cells are exactly equal, nothing happens at all. The e.m.f. of one is identical to the e.m.f. of the other. No current flows through the wire  it might as well not be there. There is in fact no p.d. so no current flows. If, however, one has a slightly higher e.m.f. than the other (almost certain to be the case really), there will be a p.d. and some current will flow through the wire until the two e.m.f.'s are equal and there is no p.d. Are you getting the difference between p.d. and e.m.f. now?
If we go back to Figure 8 for a moment, the wire that is carrying the current is getting hotter. Since it is just a drawing, we won't worry about it, but note that the e.m.f. of the cell is being used up in heating the resistance of the wire. In due course all the energy will be gone and no more current will flow. How much energy is being used up? Well, we need a new unit for energy. Energy, remember, we have defined as the capacity to do work (in this case making a wire hot, and so heating the air and so on). We are forcing electrons through a wire that has resistance, and we know that the p.d. across the wire will be the same as the e.m.f. of the cell  in this theoretical example, 1.5 volts. A practical definition of energy is 'the rate at which work is done over a period of time'  for example, how much heat will a heater give in a minute, how far can the winch raise the anchor in a minute and so on. So to define energy, first we need the rate at which work is being done. We know that the work is being done in forcing a current through the p.d. across the wire. The units for p.d. are volts, the units for current are amps, so the electrical unit for 'rate of doing work' is the voltampere, which has its own name, the watt. Work is being done at the rate of one watt when one ampere flows against a p.d. of one volt. This leads to another simple formula: watts=volts*amps The general indicator for 'power' is the letter P so we can also write:
P=V*I
This equation can be manipulated in the same way as the first one, and this is shown in Figure 11 . There is more, however. We already have equations for two of the quantities in this formula, V=I*R and I=V/R. So we can substitute (I*R) for V and come up with
P=(I*R)*I
which gives
P=I^{2}R
and
P=V*(V/R)
giving
P=V^{2}/R
In other words, if we know any two of the quantities P, V, I or R, we can calculate the other two in various ways. So, we have a definition for 'rate at which work is done' or power and a unit, the watt. The symbol for the watt is, not surprisingly, W. You would expect there to be a direct relationship between electrical power and mechanical power, and so there is. One horsepower (imperial) is equal to 746 watts.
We have yet to deal with energy. Energy is (rate at which work is being done * length of time) so we could define energy as wattseconds, wattweeks and so on. The actual unit used is the watthour. One watthour is that energy used in working at a rate of one watt for one hour. Its symbol is Wh. The commercial 'unit' for measuring the amount of electricity you use at home is a thousand times larger, the kWh or kilowatthour.
Earlier I said that resistance is dependent on the crosssectional area of a material and its length  as long as the material is consistent throughout its volume, but most materials we work with in electricity are chosen for their consistency, so we will take that as read for now. What if we wanted to heat up something? Look at Figure 12 . We know that if we pass a current through the wire it will get warmer. If, however, we want to heat something up at a distance, we don't want all the wire to get hot, just the bit that is heating something for us. How do we arrange this? We do it by making the part of the wire that we want to get hot have a high resistance, and the rest a low resistance. We could do this by making the wire thinner where we want the heat developed, and this is done at times (a fuse works on this principle), but usually we make the heating section out of a different type of wire that has an intrinsic high resistance, whilst we use low resistance wire to connect to it. The length of high resistance wire is usually wound in a coil to concentrate the heat, as in the normal electric fire or lamp bulb. In a lamp bulb it gets really hot, literally 'white hot', and the bulb shows a large increase in its resistance, which can be twenty times more than when it is cold. The wire in an electric fire also shows this effect, but to a much lesser degree, as it is made out of a special material to avoid this. Resistance is one of the most used properties of materials in electrics and electronics, and not just for heating. Components that are designed especially to show this quality of resistance are called resistors. How can we calculate what thicknesses of wire we need? We can use the equation given above. Let's say we have some thick wire with a resistance of 0.1 ohms per metre, and some high resistance wire of 98 ohms per metre. We have a generator giving 100 volts of e.m.f. and we want to heat up some water for tea or coffee ten metres away. We will use twenty metres of the thick cable to get the power to the place we want it (ten metres there and ten metres back), and one metre of the high resistance wire to get hot. We calculate like this:
20 metres of wire at 0.1 ohms per metre = 20 * 0.1 = 2 ohms
1 metre of wire at 98 ohms per metre= 98 ohms
Resistors in line add up, so Total = 98 + 2 = 100 ohms
(yes of course I've fiddled the figures!)
Total power = V * I
We don't need to calculate I from I=V/R if we use...
Total power P = V^{2 }/ R
V = 100
V^{2 }= 10 000
10 000 / 100 = 100 watts.
How much power does the heating section get? We can work this out in various ways. We could calculate the current through the whole cable, or we could work out the voltage across each of the resistances, or we could do both, or we could do it by simple proportion. Let's try them all:
Current: I = V / R
V = 100 : R = 100 : therefore I = 1 amp.
P = V * I or P = I^{2 }/ R
I = 1 : I^{2 }= 1 : R = 98
therefore
P = 98 / 1 = 98 watts
Voltage: V=I*R
We have already worked out I as 1 amp. So the volts across the high resistance section are:
1 * 98 = 98 volts
P = V^{2 }/ R = 98 * 98 / 98 = 98 watts.
Or using P = V * I (we know both V and I now!)
P = 98 * 1 = 98 watts
By proportion:
Total power = 100 watts
Total resistance = 100 ohms
high resistance section absorbs 98 / 100ths of the power = 98 watts.
Now clearly, in this case it was simplest just to work out the current and use P=I^{2}/R to calculate the power, but each case will be different, depending on which figures you know. You will notice that all the answers agree with each other  it would be alarming if they didn't, but there are two watts missing from the total power supplied by the source, which are wasted entirely in heating up the connecting wire. In lowvoltage work (as with most boat systems), we have to be very careful of just how much power is being wasted in the wiring, as we don't have oodles of power to spare.
I'm sure you have used a chart. On a chart there are standard symbols to represent rocks of various types, buoys of different types and so on. There aren't lots of drawings of the particular rock or buoy. In the same way there is a set of symbols that are used for electrical and electronic purposes. These are pretty much internationally standardized, and are much more useful for demonstrating principles than pictures of actual components (though there will be plenty of those). Look at Figure 13 . Here you will see the symbol for a resistor, which can be either a zigzag line or a box. I much prefer the zigzag, which to me looks more like the coil of wire in a wirewound resistor, so I will be using this symbol exclusively, but you will probably see the box used in some diagrams. There is also a symbol for a simple cell or source of e.m.f., of the sort we have already discussed, and this always has the long bar representing the positive end of the cell (the one the electrons go to) so the plus indicator is often omitted. The symbol for a battery indicates several cells all connected in line, but just the first and last are shown in the symbol. There are also symbols for a potentiometer and a variable resistor, which we will explain in a minute, and a very basic circuit a bit like the heating arrangement we just discussed. In this there is a source of e.m.f. (the battery), connecting wires and a resistor to develop the p.d. Note however that in the symbolic form the wires are assumed to have no resistance at all. If we wanted to represent the heating element (or load) and the resistance of the connecting wires, it would have to be drawn as in Figure 14 . Now we have a source of e.m.f. and three resistors, two representing the resistance of the wires and one the heating element or load. The connecting wires now can represent connections with zero resistance.
Why is it done this way, when there is no such thing ordinarily available as a wire with no resistance? It is done so that we can deal with the properties of the wiring and components that interest us at the moment, ignoring factors that aren't too important  another working approximation. How many volts p.d. will we find across each of the resistors? Going by this theoretical representation, we have:
e.m.f.=100V
Total resistance = 100 ohms
I=V/R= 1 amp.
This current is flowing through all the components, so, for the two 1 ohm resistors:
V=I*R = 1*2 = 1 volt (twice)
For the 98 ohm resistor
98*1=98 volts
As a quick check, the p.d.s should obviously add up to the e.m.f.: 98+2=100 volts. This is very fine. But if we tried to make this circuit, and measured the results, what might we get?
What might happen is shown in Figure 15 . To our surprise, the voltage across the 98 ohm resistor is only 97.02 volts, not the 98 we had neatly calculated. The voltage across the 1 ohm resistors is only 0.99 volts, not 1 volt as calculated. If we add up the p.d.s we get 0.99+0.99+97.02=99 volts, not 100 as we should. Why? The reason for the difference is that we are using a battery as our source. The battery, being imperfect as we all are, does not just produce an e.m.f. It is also a conductor, otherwise the current wouldn't be able to go through it, and conductors have resistance. How much in this case?
First we need to know the current. We know the voltage across the resistors, so:
I = V / R = 99 / 100 = 0.99 amps
The drop in voltage from what we expect is 10099 = 1 volt. The unexpected resistor can now be calculated by:
R =V / I = 1 / 0.99 = 1.01 ohms.
Here we have a source of e.m.f (the battery) and as soon as a current is drawn a pd is created across the internal resistance, the pd being in opposition to the e.m.f, so the voltage is reduced. To make the diagram accurate we should draw this resistor in somewhere  I have shown it in line with the positive terminal of the battery. This resistance is, reasonably enough, known as the internal resistance of the battery. All sources of e.m.f. have an internal resistance. Often we don't have to worry about it, but it is always there. Why have I drawn it in line (the proper term is 'in series') with the positive terminal? Well, we need to choose a point to use as a reference for making measurements. So far it has been unnecessary as we have just measured across the resistors, but remember that this is just a representation of the real circuit, where the resistors are actually pieces of wire. It would be more convenient to measure from the battery terminal across the wire to the 98 ohm load in practice.
We could of course choose any point as a reference point, but the convention in this sort of circuit is to use the negative terminal of the battery as the reference and measure everything from that, as shown in Figure 16 . I will explain why a bit later. Since the negative terminal is the reference for now, the resistor representing the battery internal resistance is inserted in the positive lead so that its effects do not mess up our reference point. In most of the circuits we will be using this will be the convention, but to make it clear the reference point should be marked. Please notice that the relative measurements are preceded by a (+) sign. This is because we are measuring the potential relative to the reference point. Since we have used the most negative point for our reference, all the measurements are, in this case, more positive, but this will not be the case for all circuits.
Since all normal conductors have resistance, and all components are made out of conductors, we end up with a lot of resistances in circuits. We need to be able to calculate their effects in various combinations. Look at Figure 17 . Here we have a selection of circuits, each with a battery of 13.8 volts connected to various resistors. I show the resistances as resistors, but they could be anything that has a resistance of the value shown, the calculations would be the same. It's the resistance we're interested in. The choice of the voltage of the battery may or may not ring a bell  in fact, this is the nominal fullycharged voltage of a socalled 12 volt leadacid battery. The first circuit, Figure 17 shows two resistors in series, that is connected in line. You know how to deal with this one. What current is flowing?
I = V / R =13.8 / 3000
= 0.0046 amps.
What is the p.d across the two resistors? The same as the battery voltage, 13.8 volts. What is the p.d. across each resistor? We can either use the current that we just derived, or do it by ratios  simple proportion.
I have labelled the first two resistors R1 and R2 to avoid confusion. They are normally given numbers in this way.
R1 = 1000 ohms
I * R1 = 4.6 volts
We could if we wanted subtract this from 13.8 volts to give 9.2 volts across the second resistor, but we'll calculate it anyway:
R2 = 2000 ohms
I * R2 = 9.2 volts
By proportion: Total resistance is
R1 + R2 = 3000 ohms
Voltage across R2 will be:
13.8 * (2000 / 3000) = 13.8 * (2 / 3) = 9.2 volts
and by the same logic the voltage across R1 will be:
13.8 * (1000 / 3000) = 13.8 * (1 / 3) = 4.6 volts.
Why I am emphasizing these different methods? It is because when in practice you need to know the voltage across something, you may well not have all the components neatly marked with their values as in this diagram. You may only know one or two figures, and therefore versatility is a virtue. I have marked the voltages on the diagram as before, first as voltages across the resistors and then as voltages related to the negative terminal of the battery.
You probably can see that by varying the ratios of the values of the two resistors, we could arrange any permutation of voltage across each of them, from 0 to 13.8 volts. What is happening is that the p.d. across the two is divided in the ratio of the resistors, so this is known as a potential divider. This turns out to be such a useful function that apart from fixedvalue resistors in component form being used for exactly this job most of the time, there is also a huge range of components readymade for the job, called variable resistors or more commonly potentiometers (abbreviated to 'pots'). I mentioned these earlier, and there is one in action in Figure 17 , marked VR1. It consists of a resistor with a contact which can be moved from end to end, so that we can produce any fraction of the input voltage across the ends between either end and the moving contact. If this contact was set so that the top half was 1000 ohms and the bottom half was 2000 ohms, we would have an equivalent to the previous circuit. With regard to the negative terminal of the battery, the voltage would be +13.8 at the top of the pot and +9.2 volts at the moving contact. We have lost 4.6 volts across the 1000 ohm section, and so we can say that there is a voltage drop of 4.6 volts across the 1000 ohm part of the pot. If the sliding contact of the pot was connected by a wire to one end, we would now have a resistor that could be varied from zero to the value of the resistance  in the case of the 3000 ohm pot, from zero to 3000 ohms. This is also a useful function, but variable resistors of this type are not common components as they can be made so easily from a pot. The symbol for the variable resistor is also in Figure 13 , next to the pot [NOTE 4] .
Let's calculate the power used up in the resistors. The total power will be the same for both Figure 17a and b, so we'll start with that.
P = V * I
we know V = 13.8 volts and I = 0.0046 amps, so:
P = 13.8 * 0.0046 = 0.063 watts.
This is not going to do much to heat up the cabin. How is the power divided between the two resistors? using
P = V^{2 }/ R
 since we know V=4.6 volts and R=1000 ohms for R1 
P = 13.8 * 13.8 / 1000 = 0.021 watts
and for R2 with V = 9.2 volts and R = 2000 ohms
P = 9.2 * 9.2 / 2000 = 0.042 watts
Checking, 0.042 + 0.021 = 0.063 watts. As we would expect, the power also divides in the ratio of the resistors, two to one. I would hope you are getting the idea by now, so let's look at a different arrangement.
In Figure 18 there are two resistors connected in parallel across the battery. How much current is flowing in total? We know that the p.d across each resistor will be equal to the e.m.f of the battery, 13.8 volts, so we can calculate the current through each one. For R3:
I = V / R = 13.8 / 1000 = 0.0138 amps
For R4:
I = V / R = 13.8 / 2000 = 0.0069 amps
The total current is then 0.0207 amps.
Now if we calculate what resistance would allow a current of this amount to flow from R = V / I:
R = 13.8 / 0.0207 = 666.66 ohms.
Now this value is lower than either of the two resistors. They are sharing the current between them in the ratio of their resistances, and acting as one smaller value resistor. We will look at a couple more examples to get the feel of this. In Figure 19 there is a similar circuit. This time the two resistors have the same value, 1000 ohms. We'll try the figures again:
Current for R5:
I = V / R = 13.8/1000
= 0.0138 amps
Current for R6:
I = V / R = 13.8 / 1000
= 0.0138 amps
Total current is:
0.0138 + 0.0138 = 0.0276 amps
This is equivalent to a resistance of:
R = 13.8 / 0.0276 = 500 ohms.
The two resistors of equal value in parallel act like one resistor of half the value. Obviously this will always be true for any two equal resistances in parallel. Now look at Figure 20 . Here we have another parallel network, this time with one resistor of 1000 ohms and one of 1,000,000 ohms. The calculations are:
Current for R7 is as for before, I = 0.0138 amps
Current for R8:
I = V / R = 13.8 / 1,000,000 = 0.0000138 amps
Total current = 0.0138138 amps
This looks like a resistance of:
R = 13.8 / 0.138138 = 999 ohms
The very large resistance has practically no effect here. The resistors have an effect in inverse or reciprocal proportion to their size. We need a general solution for this case of resistors in parallel.
You can see that any number of resistors in parallel will act as one resistor of a lower value. For any resistor, the current through it is I=V/R. The total current is the sum of the individual currents.
We can say for two resistors Ra and Rb:
I total =V / Ra + V/ Rb
but I total is equal to V / (the equivalent resistor) or (R_{EQ}),
so we can say:
V / R_{EQ} = V / Ra + V / Rb
Since they all have the same V from the same battery at the top, it can be taken out, giving:
1 / R_{EQ} = 1 / Ra + 1 / Rb
This sort of equation with 1 divided by something is called a reciprocal. In the general case, for any number of resistors in parallel R1 to Rn, we get:
For total resistance Rt: 1 / Rt = 1 / R1 + 1 / R2 + ... 1 / Rn where Rt is the result, and R1, R2... Rn are the various resistors.
Now this isn't at all difficult, but you can probably see why electrics people tend to buy calculators. An example:
Two resistors in parallel, each of 1000 ohms (just to check it out)
1 / Rt = 1 / 1000 + 1 / 1000 = 0.001 + 0.001 = 0.002
Therefore
Rt = 1 / (0.002) = 500 ohms
so it works. A more complex example:
Three resistors in parallel, one of 2700 ohms, one of 3300 ohms and one of 680 ohms:
1 / Rt = 1 / R1 + 1 / R2 + 1 / R3 = 1 / 2700 + 1 / 3300 + 1 / 680 = 0.00037 + 0.000303 + 0.00147 = 0.0021439
Rt = 1 / 0.0021439 = 466.42 ohms
This is not at all tricky mathematics in my book, but I'd rather use the calculator, as I said. Because this equation does not give rise to easy longhand arithmetic, there is another one to deal with the special case of just two resistors in parallel, which does not need reciprocals. It is:
Rt = R1 * R2 / R1 + R2
This is much easier. Checking again with our known result:
Rt = (1000 * 1000) / (1000 + 1000) = 1 000 000 / 2000 = 500 ohms
So it works. For two resistors in parallel of 4700 and 10000 ohms:
Rt = (4700 * 10000) / (4700 + 10000) = 47 000 000 / 14700 = 3197 ohms
Which doesn't need a calculator. Well I do, but you may not. I do hope you are not thinking "Good Lord, not mathematics!!" Particularly as this is it for fundamental resistance, volts, amps and watts. You now know it. If you can handle the peculiarities of tidal calculations around the British Isles you can certainly manage the maths we will use. If you are flummoxed by this, the next section is a quick overview of the maths needed. We will just do a couple more interesting examples and then on to the next topic. Of course, there may be a couple more details later...
Try this calculation. You want to know what voltage reaches your starting motor when you press the button. After much research you manage to collect these figures:
These aren't far from reality, by the way. First we draw a circuit. Always this is the first thing to do when trying to solve a problem, so I will emphasise it. FIRST DRAW A CIRCUIT. I've done it for you in Figure 21 . We will add up the resistances and calculate the voltages. The calculations will be approximate at this stage, because there is more to know about motors, but nonetheless worthwhile.
Total resistance = 0.001 + 0.02=0.021 ohms.
Current is 250 amps, so volts dropped across total resistance
V = I * R = 250 * 0.021 = 5.25 volts.
This means that the total volts available for the motor are
13.8  5.25 = 8.55 volts.
Interesting, isn't it? Where have the volts gone?
Obviously most of the drop is caused by the wiring resistance of 0.02 ohms, which is a bit too high, although 0.02 ohms is a very small resistance. The drop it causes is
V = I * R = 250 * 0.02 = 5 volts.
The drop caused by the internal resistance of the battery is only 0.25 volts. Engine manufacturers know that there will be a voltage drop in the system, so the starter motor is usually designed to run on about 10 volts, which is all it can expect to get at switchon, when it takes a very large current. How much power does the starter motor take at switchon?
P = V * I = 8.55 (this is what it is getting) *250 = 2138 watts
How many horsepower?
1 horsepower = 746 watts so
2138 / 746 = 2.87hp.
If it takes 30 seconds of churning to start the engine, how many amperehours of current have been used? Well, 30 seconds is 60/0.5= 120th of an hour. If the amperes stay constant at 250A:
250 / 120 = 2.08 Ah. How many watthours of energy? We already calculated the watts as 2138, so:
2138 / 120 = 17.8 Wh.
Starter motors are discussed in more detail elsewhere in the text, but for now let's see what happens if we have a slightly dodgy connection in the wiring  nothing too terrible, just another 0.02 ohms. The figures:
R = 0.001 + 0.04 = 0.041 ohms
V = I * R = 250 * 0.041 = 10.25 volts.
Voltage reaching motor 
13.8  10.25 = 3.55 volts !!!
How many watts?
P = V * I = 888 watts
Horsepower?
888 / 746 = 1.2 hp
Note that the starter motor is now applying less than half as much power to start the engine! This is the result of the drop in the wiring, 10.25 volts, if the starter motor was taking 250 amps. In fact, on such a low voltage it would try and take much more (we will see why later), with the likely result that either it would burn out or the wiring would smoke or something else unpleasant would happen. What wouldn't happen is any sign of life from the engine. And all this because of 0.02 ohms extra resistance in the starting circuit. Perhaps you would like to check your connections? The total resistance in the starting circuit should be less than 0.015 ohms for good results. You can probably also see why a slightly flat battery + slightly dirty connections=big trouble.
The basic equation for this is
from which:
V being in volts, I in amps and R in ohms. Power is measured in watts. The equation for this is:
In conjunction with the previous equation, we can also come up with:
with P in watts, V in volts, I in amps and R in ohms.
Resistances in combination behave like this:
Resistors in series add.
Resistors in parallel of different values need a simple formula:
for any number of resistors: For total resistanceRt: 1 / Rt = 1 / R1 + 1 / R2 +... 1 / Rn
for two resistors only: Rt = R1 * R2 / R1 + R2
the value of any number of equal resistors in parallel can be calculated by simply dividing the value of one of them by the number of resistors in parallel as follows:
 four 1000 ohms in parallel  value = 1000 / 4 = 250 ohms
 six 300 ohms in parallel  value = 300 / 6 = 50 ohms
[Note 1] The discharge of the electron deficiency is only true as I have described it if the two charged surfaces are conductors, but the idea is I hope clear
[Note 2] If you did any science at school and finished before the midseventies you may well find that you have never heard of some of the units used now. This is because there was a major shakeup involving the scrapping of the centimetregramsecond (cgs) system and the introduction of the metrekilogramsecond (MKS) system, on the grounds of rationalization. This meant that various fundamental quantities were redefined. If you're from the US, I'm sorry, you'll just have to get used to the units, we did!
[Note 3] For quite some time the ampere was officially defined as 'that current which, if flowing in two straight parallel wires of infinite length, placed one metre apart in a vacuum, will produce on each of the wires a force of 2*10^{7} newtons per unit length'. If you find this a bit fishy (I haven't encountered many infinite straight wires in vacuum) join the club. Perhaps this is how some scientists have gained an impractical aura
[Note 4] The name potentiometer is really a misnomer, but is so commonplace that you might as well use it (a potentiometer is properly speaking a sort of meter for measuring potential using variable resistors). Variable resistor is better, but is usually taken to mean a component with only two terminals, that is a pot with a wire shorting one end to the slider, as described.
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