Waves: Introduce With Trolleys, Not Slinkys

A number of interesting questions came up in the recent author Q&A run by @Chatphysics about my contribution to Cracking Key Concepts in Secondary Science: many thanks to Jinny Bell (@MissB0107) for hosting!

In this post, the question I want to address in more detail is: why do I recommend introducing waves using dynamics trolleys rather than a slinky spring or elastic rope?

I think I have a couple of persuasive reasons; but first a digression.

A thing doing a thing…

Robert Graves wrote his charming comedic poem Welsh Incident in 1929. It describes (at least as I read it) the visitation of disparate group of space aliens to a small town on the North Wales coast in the 1920s.

I will paraphrase one line where the Narrator says:

Then one of things did a thing that was finally recognisable as a thing.

The first example of a thing doing a thing…

Now don’t get me wrong: slinky springs and elastic ropes are brilliant examples of mechanical waves and I do use them (a lot!) — I just don’t use them as the first example.

The problem as I see it is that they could lead students to file wave mechanics in an entirely separate and independent category from mechanics.

I want students to see that wave behaviour isn’t distinct from forces and particles but rather is a direct (and fairly straightforward) consequence of a particular arrangement of particles with a specific pattern of forces between them.

Since the first example, is often the ‘loudest’ (metaphorically speaking), it’s not a bad idea to start with longitudinal waves.

I use standard wooden dynamics trolleys. Dowel rods or metal posts can be used to link the trolleys together. The system is more stable if a pair of springs is used at the front and back of each trolley. The springs used are the ones we typically use for the Hooke’s Law experiment.

A compression carrying energy along a line of trolleys linked by springs can be easily modelled:

Modelling energy transfer by compression of a longitudinal wave using a line of dynamics trolleys and springs

So can a rarefaction:

Using a line of dynamics trolley linked together with spring to demonstrate the transmission of a rarefaction pulse

Transverse waves can be modelled like this:

A transverse wave modelled using a line of dynamics trolleys linked with springs

Amongst the advantages of this approach are:

  • Students are introduced to an unknown thing (wave behaviour) by means of more familiar things (trolleys and springs)
  • The idea that there is no net movement of the ‘particles’ as energy is transferred is much more directly observable using this arrangement rather than the slinky or elastic rope.
  • The frequency of a wave (which in some ways is a more fundamental measurement than wavelength) can be associated with the repeating motion of a single ‘particle’ and extended outwards to the whole system, rather than vice versa.

You can read more in Chapter 25 of Cracking Key Concepts in Secondary Science.

Conclusion

I hope readers will try this demonstation: hopefully introducing students to a thing which is already recognisable as a thing will make wave behaviour more comprehensible and less like an unwelcome diversion into terra incognita.

Readers who are ‘rich in years’ like myself will recognise this demonstration as being adapted from the old Nuffield linear A-level Physics course.

You can listen to Richard Burton’s great reading of Robert Graves’ Welsh Incident here.

Circuit Diagrams: Lost in Rotation…?

Is there a better way of presenting circuit diagrams to our students that will aid their understanding of potential difference?

I think that, possibly, there may be.

(Note: circuit diagrams produced using the excellent — and free! — web editor at https://www.circuit-diagram.org/.)

Old ways are the best ways…? (Spoiler: not always)

This is a very typical, conventional way of showing a simple circuit.

A simple circuit as usually presented

Now let’s measure the potential difference across the cell…

Measuring the potential difference across the cell

…and across the resistor.

Measuring the potential difference across the resistor

Using a standard school laboratory digital voltmeter and assuming a cell of emf 1.5 V and negligible internal resistance we would get a value of +1.5 volts for both positions.

Let me demonstrate this using the excellent — and free! — pHET circuit simulation website.

Indeed, one might argue with some very sound justification that both measurements are actually of the same potential difference and that there is no real difference between what we chose to call ‘the potential difference across the cell’ and ‘the potential difference across the resistor’.

Try another way…

But let’s consider drawing the circuit a different (but operationally identical) way:

The same circuit drawn ‘all-in-a-row’

What would happen if we measured the potential difference across the cell and the resistor as before…

This time, we get a reading (same assumptions as before) of [positive] +1.5 volts of potential difference for the potential difference across the cell and [negative] -1.5 volts for the potential difference across the resistor.

This, at least to me, is a far more conceptually helpful result for student understanding. It implies that the charge carriers are gaining energy as they pass through the cell, but losing energy as they pass through the resistor.

Using the Coulomb Train Model of circuit behaviour, this could be shown like this:

+1.5 V of potential difference represented using the Coulomb Train Model
-1.5 V of potential difference represented using the Coulomb Train Model. (Note: for a single resistor circuit, the emerging coulomb would have zero energy.)

We can, of course, obtain a similar result for the conventional layout, but only at the cost of ‘crossing the leads’ — a sin as heinous as ‘crossing the beams’ for some students (assuming they have seen the original Ghostbusters movie).

Crossing the leads on a voltmeter

A Hidden Rotation?

The argument I am making is that the conventional way of drawing simple circuits involves an implicit and hidden rotation of 180 degrees in terms of which end of the resistor is at a more positive potential.

A hidden rotation…?

Of course, experienced physics learners and instructors take this ‘hidden rotation’ in their stride. It is an example of the ‘curse of knowledge’: because we feel that it is not confusing we fail to anticipate that novice learners could find it confusing. Wherever possible, we should seek to make whatever is implicit as explicit as we can.

Conclusion

A translation is, of course, a sliding transformation, rather than a circumrotation. Hence, I had to dispense with this post’s original title of ‘Circuit Diagrams: Lost in Translation’.

However, I do genuinely feel that some students understanding of circuits could be inadvertently ‘lost in rotation’ as argued above.

I hope my fellow physics teachers try introducing potential difference using the ‘all-in-row’ orientation shown.

The all-in=a-row orientation for circuit diagrams to help student understanding of potential difference

I would be fascinated to know if they feel its a helpful contribition to their teaching repetoire!

Booklet for teaching the Coulomb Train Model

At the bottom of the post are some links to a student booklet for teaching part of the electricity content for AQA GCSE Physics / AQA GCSE Combined Science using the Coulomb Train Model.

Extract from booklet

I have believed for a long time that the electricity content is often ‘under-explained’ at GCSE: in other words, not all of the content is explicitly taught. I have deliberately have gone to the opposite extreme here — indeed, some teachers may feel that I have ‘over-explained’ too much of the content. However, the booklets are editable so feel free to adapt!

I think the booklet is suitable for teacher-led instruction as well as independent study — I would love to hear how your students have responded to it.

The animations will be ‘live’ for the Google Docs and MS Word versions, but will be frozen for the PDF version. They can be cut and pasted into Powerpoint or other teaching packages (but please note that in some versions of PPT, the animations will appear frozen until you go into presenter mode).

Please feel free to download, use and adapt as you see fit. It is released under the terms of the Creative Commons Attribution License CC BY-SA 4.0 (details here), so please flag if you see versions being sold on TES or similar websites.

The remaining content for AQA electricity will be released (fingers crossed) over the next couple of months.

Feedback and comments (hopefully mainly positive) always welcome….

Visualising How Transformers Work

‘Transformers’ is one of the trickier topics to teach for GCSE Physics and GCSE Combined Science.

I am not going to dive into the scientific principles underlying electromagnetic induction here (although you could read this post if you wanted to), but just give a brief overview suitable for a GCSE-level understanding of:

  • The basic principle of a transformer; and
  • How step down and step up transformers work.

One of the PowerPoints I have used for teaching transformers is here. This is best viewed in presenter mode to access the animations.

The basic principle of a transformer

A GIF showing the basic principle of a transformer.
(BTW This can be copied and pasted into a presentation if you wish,)

The primary and secondary coils of a transformer are electrically isolated from each other. There is no charge flow between them.

The coils are also electrically isolated from the core that links them. The material of the core — iron — is chosen not for its electrical properties but rather for its magnetic properties. Iron is roughly 100 times more permeable (or transparent) to magnetic fields than air.

The coils of a transformer are linked, but they are linked magnetically rather than electrically. This is most noticeable when alternating current is supplied to the primary coil (green on the diagram above).

The current flowing in the primary coil sets up a magnetic field as shown by the purple lines on the diagram. Since the current is an alternating current it periodically changes size and direction 50 times per second (in the UK at least; other countries may use different frequencies). This means that the magnetic field also changes size and direction at a frequency of 50 hertz.

The magnetic field lines from the primary coil periodically intersect the secondary coil (red on the diagram). This changes the magnetic flux through the secondary coil and produces an alternating potential difference across its ends. This effect is called electromagnetic induction and was discovered by Michael Faraday in 1831.

Energy is transmitted — magnetically, not electrically — from the primary coil to the secondary coil.

As a matter of fact, a transformer core is carefully engineered so to limit the flow of electrical current. The changing magnetic field can induce circular patterns of current flow (called eddy currents) within the material of the core. These are usually bad news as they heat up the core and make the transformer less efficient. (Eddy currents are good news, however, when they are created in the base of a saucepan on an induction hob.)

Stepping Down

One of the great things about transformers is that they can transform any alternating potential difference. For example, a step down transformer will reduce the potential difference.

A GIF showing the basic principle of a step down transformer.
(BTW This can be copied and pasted into a presentation if you wish,)

The secondary coil (red) has half the number of turns of the primary coil (green). This halves the amount of electromagnetic induction happening which produces a reduced output voltage: you put in 10 V but get out 5 V.

And why would you want to do this? One reason might be to step down the potential difference to a safer level. The output potential difference can be adjusted by altering the ratio of secondary turns to primary turns.

One other reason might be to boost the current output: for a perfectly efficient transformer (a reasonable assumption as their efficiencies are typically 90% or better) the output power will equal the input power. We can calculate this using the familiar P=VI formula (you can call this the ‘pervy equation’ if you wish to make it more memorable for your students).

Thus: Vp Ip = Vs Is so if Vs is reduced then Is must be increased. This is a consequence of the Principle of Conservation of Energy.

Stepping up

A GIF showing the basic principle of a step up transformer.
(BTW This can be copied and pasted into a presentation if you wish,)

There are more turns on the secondary coil (red) than the primary (green) for a step up transformer. This means that there is an increased amount of electromagnetic induction at the secondary leading to an increased output potential difference.

Remember that the universe rarely gives us something for nothing as a result of that damned inconvenient Principle of Conservation of Energy. Since Vp Ip = Vs Is so if the output Vs is increased then Is must be reduced.

If the potential difference is stepped up then the current is stepped down, and vice versa.

Last nail in the coffin of the formula triangle…

Although many have tried, you cannot construct a formula triangle to help students with transformer calculations.

Now is your chance to introduce students to a far more sensible and versatile procedure like FIFA (more details on the PowerPoint linked to above)

The Coulomb Train Model Revisited (Part 5)

In this post, we are going to look at series circuits using the Coulomb Train Model.

The Coulomb Train Model (CTM) is a helpful model for both explaining and predicting the behaviour of real electric circuits which I think is useful for KS3 and KS4 students.

Without further ado, here is a a summary.


A circuit with one resistor

Let’s look at a very simple circuit to begin with:

This can be represented on the CTM like this:

The ammeter counts 5 coulombs passing every 10 seconds, so the current I = charge flow Q / time t = 5 coulombs / 10 seconds = 0.5 amperes.

We assume that the cell has a potential difference of 1.5 V so there is a potential difference of 1.5 V across the resistor R1 (that is to say, each coulomb loses 1.5 J of energy as it passes through R1).

The resistor R1 = potential difference V / current I = 1.5 / 0.5 = 3.0 ohms.


A circuit with two resistors in series

Now let’s add a second identical resistor R2 into the circuit.

This can be shown using the CTM like this:

Notice that the current in this example is smaller than in the first circuit; that is to say, fewer coulombs go through the ammeter in the same time. This is because we have added a second resistor and remember that resistance is a property that reduces the current. (Try and avoid talking about a high resistance ‘slowing down’ the current because in many instances such as two conductors in parallel a high current can be modelled with no change in the speed of the coulombs.)

Notice also that the voltmeter is making identical measurements on both the circuit diagram and the CTM animation. It is measuring the total energy change of the coulombs as they pass through both R1 and R2.

The current I = charge flow Q / time t = 5 coulombs / 20 seconds = 0.25 amps. This is half the value of the current in the first circuit.

We have an identical cell of potential difference 1.5 V the voltmeter would measure 1.5 V. We can calculate the total resistance using R = V / I = 1.5 / 0.25 = 6.0 ohms.

This is to be expected since the total resistance R = R1 + R2 and R1 = 3.0 ohms and R2 = 3.0 ohms.


Looking at the resistors individually

The above circuit can be represented using the CTM as follows:

Between A and B, the coulombs are each gaining 1.5 joules since the cell has a potential difference of 1.5 V. (Remember that V = E energy transferred (in joules) / Q charge flow (in coulombs.)

Between B and C the coulombs lose no energy; that is to say, we are assuming that the connecting wires have negligible resistance.

Between C and D the coulombs lose some energy. We can use the familar V = I x R to calculate how much energy is lost from each coulomb, since we know that R1 is 3.0 ohms and I is 0.25 amperes (see previous section).

V = I x R = 0.25 x 3.0 = 0.75 volts.

That is to say, 0.75 joules are removed from each coulomb as they pass through R1 which means that (since 1.5 joules were added to each coulomb by the cell) that 0.75 joules are left in each coulomb.

The coulombs do not lose any energy travelling between D and E because, again, we are assuming negligible resistance in the connecting wire.

0.75 joules is removed from each coulomb between E and F making the potential difference across R2 to be 0.75 volts.

Thus we find that the familiar V = V1 + V2 is a direct consequence of the Principle of Conservation of Energy.


FAQ: ‘How do the coulombs know to drop off only half their energy in R1?’

Simple answer: they don’t.

This may be a valid objection for some donation models of electric circuits (such as the pizza delivery van model) but it doesn’t apply to the CTM because it is a continuous chain model (with the caveat that the CTM applies only to ‘steady state’ circuits where the current is constant).

Let’s look at a numerical argument to support this:

  • The magnitude of the current is controlled by only two factors: the potential difference of the cell and the total resistance of the circuit.
  • In other words, if we increased the value of R1 to (say) 4 ohms and reduced the value of R2 to 2 ohms so that the total resistance was still 6 ohms, the current would still be 0.25 amps.
  • However, in this case the energy dissipated by each coulomb passing through R1 would V = I x R = 0.25 x 4 = 1 volt (or 1 joule per coulomb) and similarly the potential difference across R2 would now be 0.5 volts.
  • The coulombs do not ‘know’ to drop off 1 joule at R1 and 0.5 joules at R2: rather, it is a purely mechanical interaction between the moving coulombs and each resistor.
  • R1 has a bigger proportion of the total resistance of the circuit than R2 so it seems self-evident (at least to me) that the coulombs will lose a larger proportion of their total energy passing through R1.
  • A similar analysis would apply if we made R2 = 4 ohms and R1 = 2 ohms: the coulombs would now lose 0.5 joules passing through R1 and 1 joule passing through R2.

Thus, we see that the current in a series circuit is affected by the ‘global’ or ‘whole circuit’ properties such as the potential difference of the cell and the total resistance of the circuit. The CTM models this property of real circuits by being a continuous chain of mechanically-linked ‘trucks’ so that a change in any one part of the circuit affects the movement of all the coulombs.

However, the proportion of the energy lost by a coulomb travelling through one part of the circuit is affected — not by ‘magic’ or a weird form of ‘coulomb telepathy’ — but only by the ‘local’ properties of that section of the circuit i.e. the electrical resistance of that section.

The CTM analogue of a low resistance section of a circuit (top) and a high resistance section of a circuit (bottom)

(PS You can read more about the CTM and potential divider circuits here.)


Afterword

You may be relieved to hear that this is the last post in my series on ‘The CTM revisited’. My thanks to the readers who have stayed with me through the series (!)

I will close by saying that I have appreciated both the expressions of enthusiasm about CTM and the thoughtful criticisms of it.

Mnemonics for the S.I. Prefixes

The S.I. System of Weights and Measures may be a bit of a dog’s dinner, but at least it’s a dog’s dinner prepped, cooked, served and — more to the point — eaten by scientists.

A brief history of the Système international d’unités

It all began with the métre (“measure”), of course. This was first proposed as a universal measure of distance by the post-Revolutionary French Academy of Sciences in 1791. According to legend (well, not legend precisely — think of it as random speculative gossip, if you prefer), they first proposed that the metre should be one millionth of the distance from the North Pole to the equator.

When that turned out to be a little on the large side, they reputedly shrugged in that inimitable Gallic fashion said: “D’accord, faisons un dix millionième alors, mais c’est ma dernière offre.” (“OK, let’s make it one ten millionths then, but that’s my final offer.”)

Since then, what measurement-barbarians loosely (and egregiously incorrectly) call “the metric system” has been through many iterations and revisions to become the S.I. System. Its full name is the Système international d’unités which pays due honour to France’s pivotal role in developing and sustaining it.

When some of those same measurement-barbarians call for a return to the good old “pragmatic” Britsh Imperial System of inches and poundals, I urge all fair-minded people to tell them, as kindly as possible, that they can’t: not now, not ever.

Since 1930, the inch has been defined as 25.4 millimetres. (It was, so I believe, the accuracy and precision needed to design and build jet engines that led to the redefinition. The older definitions of the inch simply weren’t precise enough.)

You simply cannot replace the S.I. system, you can, however, dress it up a little bit and call a distance of 25.4 millimetres “one inch” if you really wanted to — but, in the end, what would be the point of that?

The Power of Three (well, ten to the third power, anyways)

For human convenience, the S.I. system includes prefixes. So a large distance might measured in kilometres where the prefix kilo- indicates multiplying by a factor of 1000 (or 10 raised to the third power). The distance between the Globe Theatre in London and Slough Station is 38.6 km. Longer distances such as London and New York, NY would be 5.6 megametres (or 5.6 Mm — note capital ‘M’ for mega [one million] to avoid confusion with the prefix milli- ).

The S.I. System has prefixes for all occasions, as shown below.

The ‘big’ SI prefixes.
Note that every one of them, except for kilo, is represented by a capital letter.

Note also that one should convert all prefixes into standard units for calculations e.g. meganewtons should be converted to newtons. The sole exception is kilograms because the base unit is the kilogram not the gram, so a megagram should be converted into kilograms, not grams. I trust that’s clear. (Did I mention the “dog’s dinner” part yet?)

For perspective, the distance between Earth and the nearest star outside our Solar System is 40 petametres, and current age of the universe is estimated to be 0.4 exaseconds (give or take a petasecond or two).

A useful mnemonic for remembering these is Karl Marx Gives The Proletariat Eleven Zeppelins (and one can imagine the proletariat expressing their gratitude by chanting in chorus: “Yo! Ta, Mr Marx!” as they march bravely forward.)

Karl Marx Gives The Proletariat Eleven Zeppelins (“Yo! Ta, Mr. Marx!)

But what about the little prefixes?

Milli- we have already covered above. The diameter of one of your red blood cells in 8 micrometres and the time it takes light to travel a distance equal to the diameter of a hydrogen atom is 300 zeptoseconds.

Again, there is an SI prefix for every occasion:

The ‘little’ SI prefixes.
(Handily, all of them are represented by lower case letters — including micro which is shown the lower case Greek letter ‘mu’)

A useful mnemonic would be: Millie’s Microphone Needs a Platform For Auditioning Zebra Yodellers.

For the record, GCSE Physics students are expected to know the SI prefixes between giga- and pico-, but if you’re in for a pico- then you’re in for anything between a yotta- and a yocto- in my opinion (if you catch my drift).

Very, very, very small to very, very, very big

The mean lifetime of a Z-boson (the particle that carries the Weak force) is 0.26 yoctoseconds.

According to our current understanding of physics, the stars will have stopped shining and all the galaxies will dissipate into dissassociated ions some 315 yottaseconds from now.

Apart from that, happy holidays everyone!

FIFA and Really Challenging GCSE Physics Calculations

‘FIFA’ in this context has nothing to do with football; rather, it is a mnemonic that helps KS3 and KS4 students from across the attainment range engage productively with calculation questions.

FIFA stands for:

  • Formula
  • Insert values
  • Fine-tune
  • Answer

From personal experience, I can say that FIFA has worked to boost physics outcomes in the schools I have worked in. What is especially gratifying, however, is that a number of fellow teaching professionals have been kind enough to share their experience of using it:


Framing FIFA as a modular approach

Straightforward calculation questions (typically 2 or 3 marks) can be ‘unlocked’ using the original FIFA approach. More challenging questions (typically 4 or 5 marks) can often be handled using the FIFA-one-two approach.

However, what about the most challenging 5 or 6 mark questions that are targeted at Grade 8/9? Can FIFA help in solving these?

I believe it can. But before we dive into that, let’s look at a more traditional, non-FIFA, algebraic approach.


A challenging freezing question: the traditional (non-FIFA) algebraic approach

Note: this is a ‘made up’ question written in the style of the GCSE exam.

A pdf of this question is here. A traditional algebraic approach to solving this problem would look like this:

This approach would be fine for confident students with high previous attainment in physics and mathematics. I will go further and say that it should be positively encouraged for students who possess — in Edward Gibbon’s words — that ‘happy disposition’:

But the power of instruction is seldom of much efficacy, except in those happy dispositions where it is almost superfluous.

Edward Gibbon, The Decline and Fall of the Roman Empire

But what about those students who are more akin to the rest of us, and for whom the ‘power of instruction’ is not a superfluity but rather a necessity on which they depend?


A challenging freezing question: the FIFA-1-2-3 approach

Since this question involves both cooling and freezing it seems reasonable to start with the specific heat capacity formula and then use the specific latent heat formula:

FIFA-one-two isn’t enough. We must resort to FIFA-1-2-3.

What is noteworthy here is that the third FIFA formula isn’t on the formula sheet and is not on the list of formulas that need to be memorised. Instead, it is made by the student based on their understanding of physics and a close reading of the question.

Challenging? Yes, undoubtedly. But students will have unlocked some marks (up to 4 out of 6 by my estimation).

FIFA isn’t a royal road to mathematical mastery (although it certainly is a better bet than the dreaded ‘formula triangle’ that I and many other have used in the past). FIFA is the scaffolding, not the finished product.

Genuine scientific understanding is the clock tower; FIFA is simply some temporary scaffolding that helps students get there.

We complete the FIFA-1-2-3 process as follows:


Conclusion: FIFA fixes it

The FIFA-system was born of the despair engendered when you mark a set of mock exam papers and the majority of pages are blank: students had not even attempted the calculation skills.

In my experience, FIFA fixes that — students are much more willing to start a calculation question. And that means that, even when they cannot successfully navigate to a ‘full mark’ conclusion, they gain at least some marks, and and one does not have to be a particularly perceptive scholar of the human heart to understand that gaining ‘some marks‘ is more motivating than ‘no marks‘.

The Coulomb Train Model Revisited (Part 3)

In this post, we will look at explaining electrical resistance using the Coulomb Train Model.

This is part 3 of a continuing series (click to read part 1 and part 2).

The Coulomb Train Model (CTM) is a helpful model for both explaining and predicting the behaviour of real electric circuits which I think is useful for KS3 and KS4 students.

Without further ado, here is a a summary.

A summary of the Coulomb Train Model

Representing Resistance on the CTM

To measure resistance, we would set up this circuit.

We can represent this same circuit on the CTM as follows:

If we count how many ‘coulombs’ (grey trucks) pass one point in a certain time then on this animation we get 5 coulombs in 20 seconds.
This is equivalent to a current of
5 coulombs / 20 seconds = 0.2 coulombs per second = 0.2 amperes.

This way of thinking about current is consistent with the formula charge flow = current x time or Q=It which can be rearranged to give I=Q/t.

We have used identical labels on the circuit diagram and the CTM animation to encourage students to view them as different representations of a real situation. The ammeter at X would read 0.2 amps. We could place the ammeter at any other point in the circuit and still get a reading of 0.2 amps since ammeters only ‘count coulombs per second’ and don’t make any measurement of energy (represented by the orange substance in the trucks).

However, the voltmeter does make a measurement of energy: it compares the energy difference between a single coulomb at Y and a single coulomb at Z. If (say) 1.5 joules of energy is transferred from each coulomb as it passes through the bulb from Y to Z then the voltmeter will read a potential difference (or ‘voltage’ if you prefer) of 1.5 volts.

This way of thinking about potential difference is consistent with the formula energy transferred = charge flow x potential difference or E=QV which we can rearrange to give V=E/Q.

So as you can see, one volt is really equivalent to an energy change of one joule for every coulomb (!)

We can calculate the resistance of the bulb by using R=V/I so R = 1.5/0.2 = 7.5 ohms.


Resistance is not futile . . .

Students sometimes have difficulty accepting the idea of a ‘resistor’: ‘Why would anyone in their right minds deliberately design something that reduces the flow of electric current?’ It’s important to explain that it is vital to be able to control the flow of electric current and that one of the most common electronic components in your phone or games console is — the humble resistor.

One of many resistors on a circuit board. The colour codes tell us the value of each resistor.

Teachers often default to explaining electric circuits using bulbs as the active component. There is a lot to recommend this practice, not least the fact that changes in the circuit instantaneously affect the brightness of the bulb. However, it vital (especially at GCSE) to allow students to learn about circuits featuring resistors and other components rather than just the pedagogically overused (imho) filament lamp.


Calculating the resistance of a resistor

Consider this circuit where we have a resistor R1.

This can be represented as a coulomb train model like this:

The resistor does not glow with visible light as the bulb does, but it would glow pretty brightly if viewed through an infra red camera since the energy carried by the coulombs is transferred to the thermal energy store of the resistor. The only way we can observe this energy shift without such a special camera is to use a voltmeter.

Let’s begin by analysing this circuit qualitatively.

  • The coulombs are moving faster in this circuit than the previous circuit. This means that the current is larger. (Remember: current is coulombs per second.)
  • Because the current is larger, R1 must have a smaller resistance than the bulb. (Remember: resistance is a quantity that reduces the current.)
  • The energy transferred to each coulomb is the same in each example so the potential difference of the cell is the same in both circuits. (Of course, V can be altered by adding a second cell or turning up the setting on a power supply, but in many circuits V is, loosely speaking, a ‘fixed’ or ‘quasi-constant’ value.)
  • Because the ‘push’ or potential difference is the same size but the resistance of R1 is smaller, then the same cell is able to push a larger current around the circuit.

Now let’s analyse the circuit quantitatively.

  • 5 coulombs pass a single point in 13 seconds so the current is 5/13 = 0.38 coulombs per second = 0.4 amperes. (Double the current in the bulb circuit.)
  • The resistance can be calculated using R=V/I = 1.5/0.4 = 3.75 ohms. (Half the resistance of the bulb.)
  • Each coulomb is being loaded with 1.5 J of energy as it passes through the cell. Since this is happening twice as often in the resistor circuit as the bulb circuit, the cell will ‘go flat’ or ’empty its chemical energy store’ in half the time of the bulb cell.

So there we have it: more fun and high jinks with the CTM.

I hope that I have persuaded a few more teachers that the CTM is useful for getting students to think productively and, more importantly, quantitatively using correct scientific terminology about electric circuits.

In the next installment, we will look at series and parallel circuits.

The Coulomb Train Model Revisited (Part 1)

The Coulomb Train Model (CTM) is a helpful model for both explaining and predicting the behaviour of real electric circuits which I think is suitable for use with KS3 and KS4 students (that’s 11-16 year olds for non-UK educators).

I have written about it before (see here and here) but I have recently been experimenting with animated versions of the original diagrams.

Essentially, the CTM is a donation model akin to the famous ‘bread and bakery van’ or even the ‘penguins and ski lift’ models, but to my mind it has some major advantages over these:

  • The trucks (‘coulombs’) in the CTM are linked in a continuous chain. If one ‘coulomb’ stops then they all stop. This helps students grasp why a break anywhere in a circuit will stop all current.
  • The CTM presents a simplified but still quantitatively accurate picture of otherwise abstract entities such as coulombs and energy rather than the more whimsical ‘bread van’ = ‘charge carrier’ and ‘bread’ = ‘energy’ (or ‘penguin’ = ‘charge carrier’ and ‘gpe of penguin’ = ‘energy of charge carrier’) for the other models.
  • Explanations and predictions scripted using the CTM use direct but substantially correct terminiology such as ‘One ampere is one coulomb per second’ rather than the woolier ‘current is proportional to the number of bread vans passing in one second’ or similar.

Modelling current flow using the CTM

The coulombs are the ‘trucks’ travelling clockwise in this animation. This models conventional current.

Charge flow (in coulombs) = current (in amps) x time (in seconds)

So a current of one ampere is one coulomb passing in one second. On the animation, 5 coulombs pass through the ammeter in 25 seconds so this is a current of 0.20 amps.

We have shown two ammeters to emphasise that current is conserved. That is to say, the coulombs are not ‘used up’ as they pass through the bulb.

The ammeters are shown as semi-transparent as a reminder that an ammeter is a ‘low resistance’ instrument.


Modelling ‘a source of potential difference is needed to make current flow’ using the CTM

Energy transferred (in joules) = potential difference (in volts) x charge flow (in coulombs) 

So the potential difference = energy transferred divided by the number of coulombs.

The source of potential difference is the number of joules transferred into each coulomb as it passes through the cell. If it was a 1.5 V cell then 1.5 joules of energy would be transferred into each coulomb.

This is represented as the orange stuff in the coulombs on the animation.

What is this energy? Well, it’s not ‘electrical energy’ for certain as that is not included on the IoP Energy Stores and Pathways model. In a metallic conductor, it would be the sum of the kinetic energy stores and electrical potential energy stores of 6.25 billion billion electrons that make up one coulomb of charge. The sum would be a constant value (assuming zero resistance) but it would be interchanged randomly between the kinetic and potential energy stores.

For the CTM, we can be a good deal less specific: it’s just ‘energy’ and the CTM provides a simplified, concrete picture that allows us to apply the potential difference equation in a way that is consistent with reality.

Or at least, that would be my argument.

The voltmeter is shown connected in parallel and the ‘gloves’ hint at it being a ‘high resistance’ instrument.

More will follow in part 2 (including why I decided to have the bulb flash between bright and dim in the animations).

You can read part 2 here.

Teaching Newton’s Third Law

Newton’s First and Second Laws of Motion are universal: they tell us how any set of forces will affect any object.

The “Push-Me-Pull-You” made famous by Dr Doolittle

If the forces are ‘balanced’ (dread word! — saying ‘total force is zero’ is better, I think) then the object will not accelerate: that is the essence of the First Law. If the sum of the forces is anything other than zero, then the object will accelerate; and what is more, it will accelerate at a rate that is directly proportional to the total force and inversely proportional to the mass of the object; and let’s not forget that it will also accelerate in the direction in which the total force acts. Acceleration is, after all, a vector quantity.

So far, so good. But what about the Third Law? It goes without saying, I hope, that Newton’s Third Law is also universal, but it tells us something different from the first two.

The first two tell us how forces affect objects; the third tells us how objects affect objects: in other words, how objects interact with each other.

The word ‘interact’ can be defined as ‘to act in such a way so as to affect each other’; in other words, how an action produces a reaction. However, the word ‘reaction’ has some unhelpful baggage. For example, you tap my knee (lightly!) with a hammer and my leg jerks. This is a reaction in the biological sense but not in the Newtonian sense; this type of reaction (although involuntary) requires the involvement of an active nervous system and an active muscle system. Because of this, there is a short but unavoidable time delay between the stimulus and the response.

The same is not true of a Newton Third Law reaction: the action and reaction happen simultaneously with zero time delay. The reaction is also entirely passive as the force is generated by the mere fact of the interaction and requires no active ‘participation’ from the ‘acted upon’ object.

I try to avoid the words ‘action’ and ‘reaction’ in statements of Newton’s Third Law for this reason.

If body A exerts a force on body B, then body B exerts an equal and opposite force on body A.

The best version of Newton’s Third Law (imho)

In our universe, body B simply cannot help but affect body A when body A acts on it. Newton’s Third Law is the first step towards understanding that of necessity we exist in an interconnected universe.


Getting the Third Law wrong…

Let’s consider a stationary teapot. (Why not?)

This is NOT a good illustation of Newton’s Third Law…

We can reject this as an appropriate example of Newton’s Third Law for two reasons:

  • Reason 1: Force X and Force Y are acting on a single object. Newton’s Third Law is about the forces produced by an interaction between objects and so cannot be illustrated by a single object.
  • Reason 2: Force X and Force Y are ‘equal’ only in the parochial and limited sense of being merely ‘equal in magnitude’ (8.2 N). They are very different types of force: X is an action-at-distance gravitational force and Y is an electromagnetic contact force. (‘Electromagnetic’ because contact forces are produced by electrons in atoms repelling the electrons in other atoms.) The word ‘equal’ in Newton’s Third Law does some seriously heavy lifting…

Getting the Third Law right…

The Third Law deals with the forces produced by interactions and so cannot be shown using a single diagram. Free body diagrams are the answer here (as they are in a vast range of mechanics problems).

This is a good illustration of Newton’s Third Law

The Earth (body A) pulls the teapot (body B) downwards with the force X so the teapot (body B) pulls the Earth (body A) upwards with the equal but opposite force W. They are both gravitational forces and so are both colour-coded black on the diagram because they are a ‘Newton 3 pair’.

It is worth noting that, applying Newton’s Second Law (F=ma), the downward 8.2 N would produce an acceleration of 9.8 metres per second per second on the teapot if it was allowed to fall. However, the upward 8.2 N would produce an acceleration of only 0.0000000000000000000000014 metres per second per second on the rather more massive planet Earth. Remember that the acceleration produced by the resultant force is inversely proportional to the mass of the object being accelerated.

Similarly, the Earth’s surface pushes upward on the teapot with the force Y and the teapot pushes downward on the Earth’s surface with the force Z. These two forces form a Newton 3 pair and so are colour-coded red on the diagram.

We can summarise this in the form of a table:


Testing understanding

One the best exam questions to test students’ understanding of Newton’s Third Law (at least in my opinion) can be found here. It is a really clever question from the legacy Edexcel specificiation which changed the way I thought about Newton’s Third Law because I was suddenly struck by the thought that the only force that we, as humans, have direct control over is force D on the diagram below. Yes, if D increases then B increases in tandem, but without the weighty presence of the Earth we wouldn’t be able to leap upwards…