The Coulomb Train Model Revisited (Part 4)

In this post, we will look at parallel circuits.

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.

This is part 4 of a continuing series. (Click to read Part 1, Part 2 or Part 3.)


The ‘Parallel First’ Heresy

I advocate teaching parallel circuits before teaching series circuits. This, I must confess, sometimes makes me feel like Captain Rum from Blackadder Two:

The main reason for this is that parallel circuits are conceptually easier to analyse than series circuits because you can do so using a relatively naive notion of ‘flow’ and gives students an opportunity to explore and apply the recently-introduced concept of ‘flow of charge’ in a straightforward context.

Redish and Kuo (2015: 584) argue that ‘flow’ is an example of embodied cognition in the sense that its meaning is grounded in physical experience:

The thesis of embodied cognition states that ultimately our conceptual system grounded in our interaction with the physical world: How we construe even highly abstract meaning is constrained by and is often derived from our very concrete experiences in the physical world.

Redish and Kuo (2015: 569)

As an aside, I would mention that Redish and Kuo (2015) is an enduringly fascinating paper with a wealth of insights for any teacher of physics and I would strongly recommend that everyone reads it (see link in the Reference section).


Let’s Go Parallel First — but not yet

Let’s start with a very simple circuit.

This is not a parallel circuit (yet) because switch S is open. Resistors R1 and R2 are identical.

This can be represented on the coulomb train model like this:

Five coulombs pass through the ammeter in 20 seconds so the current I = Q/t = 5/20 = 0.25 amperes.

Let’s assume we have a 1.5 V cell so 1.5 joules of energy are added to each coulomb as they pass through the cell. Let’s also assume that we have negligible resistance in the cell and the connecting wires so 1.5 joules of energy will be removed from each coulomb as they pass through the resistor. The voltmeter as shown will read 1.5 volts.

The resistance of the resistor R1 is R=V/I = 1.5/0.25 = 6.0 ohms.


Let’s Go Parallel First — for real this time.

Now let’s close switch S.

This is example of changing an example by continuous conversion which removes the need for multiple ammeters in the circuit. The changed circuit can be represented on the CTM as shown

Now, ten coulombs pass through the ammeter in twenty seconds so I = Q/t = 10/20 = 0.5 amperes (double the reading in the first circuit shown).

Questioning may be useful at this point to reinforce the ‘flow’ paradigm that we hope students will be using:

  • What will be the reading if the ammeter moved to a similar position on the other side? (0.5 amps since current is not ‘used up’.)
  • What would be the reading if the ammeter was placed just before resistor R1? (0.25 amps since only half the current goes through R1.)

To calculate the total resistance of the whole circuit we use R = V/I = 1.5/0.5 = 3.0 ohms– which is half of the value of the circuit with just R1. Adding resistors in parallel has the surprising result of reducing the total resistance of the circuit.

This is a concrete example which helps students understand the concept of resistance as a property which reduces current: the current is larger when a second resistor is added so the total resistance must be smaller. Students often struggle with the idea of inverse relationships (i.e. as x increases y decreases and vice versa) so this is a point well worth emphasising.


Potential Difference and Parallel Circuits (1)

Let’s expand on the primitive ‘flow’ model we have been using until now and adapt the circuit a little bit.

This can be represented on the CTM like this:

Each coulomb passing through R2 loses 1.5 joules of energy so the voltmeter would read 1.5 volts.

One other point worth making is that the resistance of R2 (and R1) individually is still R = V/I = 1.5/0.25 = 6.0 ohms: it is only the combined effect of R1 and R2 together in parallel that reduces the total resistance of the circuit.


Potential Difference and Parallel Circuits (2)

Let’s have one last look at a different aspect of this circuit.

This can be represented on the CTM like this:

Each coulomb passing through the cell from X to Y gains 1.5 joules of energy, so the voltmeter would read 1.5 volts.

However, since we have twice the number of coulombs passing through the cell as when switch S is open, then the cell has to load twice as many coulombs with 1.5 joules in the same time.

This means that, although the potential difference is still 1.5 volts, the cell is working twice as hard.

The result of this is that the cell’s chemical energy store will be depleted more quickly when switch S is closed: parallel circuits will make cells go ‘flat’ in a much shorter time compared with a similar series circuit.

Bulbs in parallel may shine brighter (at least in terms of total brightness rather than individual brightness) but they won’t burn for as long.

To some ways of thinking, a parallel circuit with two bulbs is very much like burning a candle at both ends…


More fun and high jinks with coulomb train model in the next instalment when we will look at series circuits.

You can read part 5 here.


Reference

Redish, E. F., & Kuo, E. (2015). Language of physics, language of math: Disciplinary culture and dynamic epistemologyScience & Education24(5), 561-590.

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 Revisited (Part 2)

In this post, we will look at understanding potential difference (or voltage) using the Coulomb Train Model.

This is part 2 of a continuing series. You can read part 1 here.

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).

To summarise what has been discussed so far:


Modelling potential difference using the CTM

Potential difference is the ‘push’ needed to make electric charge move around a closed circuit. On the CTM, we can represent the ‘push’ as a gain in the energy of the coulomb. (This is consistent with the actual definition of the volt V = E/Q, where one volt is a change in energy of one joule per coulomb.)

How can we observe this gain in energy? Simple, we use a voltmeter.

Kudos to https://www.circuit-diagram.org/editor/ for the lovely circuit diagrams

On the CTM, this would look like this:

What the voltmeter does is compare the energy contained by two coulombs: one at A and the other at B. The coulombs at B, having passed through the 1.5 V cell, each have 1.5 joules of energy more than than the coulombs at A. This means that the voltmeter in this position reads 1.5 volts. We would say that the potential difference across the cell is 1.5 V. (Try and avoid talking about the potential difference ‘through’ or ‘of’ any part of the circuit.)


More potential difference measurements using the CTM

Let’s move the voltmeter to a different position.

On the CTM, this would look like this:

Let’s make the very reasonable assumption that the connecting wires have zero resistance. This would mean that the coulombs at C have 1.5 joules of energy and that the coulombs at D have 1.5 joules of energy. They have not lost any energy since they have not passed through any part of the circuit that actually has a resistance. The voltmeter would therefore read 0 volts since it cannot detect any energy difference.

Now let’s move the voltmeter one last time.

On the CTM, this would look like this:

Notice that the coulombs at F have 1.5 fewer joules than the coulombs at E. The coulombs transfer 1.5 joules of energy to the bulb because the bulb has a resistance.

Any part of the circuit that has non-zero resistance will ‘rob’ coulombs of their energy. On this very simple model, we assume that only the bulb has a resistance and so only the bulb will ‘push back’ against the movement of the coulombs and cost them energy.

Also on this simple model, the potential difference across the bulb is identical to the potential difference across the cell — but this is not always the case. For example, if the wires had a small but non-negligible resistance and if the cell had an internal resistance, but these would only come into play at A-level.

The bulb is shown as ‘flashing’ on the CTM to provide a visual cue to help students mentally model the transfer of energy from the coulombs to the bulb. In reality, instead of just one coulomb transferring a largish ‘chunk’ of energy, there would be approximately 1.25 billion billion electrons continuously transferring a tiny fraction of this energy over the course of one second (assuming a d.c. current of 0.2 amps) so we wouldn’t see the bulb ‘flash’ in reality.


How do the coulombs ‘know’ how much energy to drop off?

This section is probably more of interest to specialist physics teachers, but all are welcome.

One frequent criticism of donation models like the CTM is how do the coulombs ‘know’ to drop off all their energy at the bulb?

The response to this, of course, is that they don’t. This criticism is an artefact of an (arguably) over-simplified model whereby we assume that only the bulb has resistance. The energy carried by the coulombs according to this model could be shown as a sketch graph, and let’s be honest it does look a little dodgy…

But, more accurately, of course, the energy loss is a process rather than an event. And the connecting wires actually have a small resistance. This leads to this graph:

Realistically speaking, the coulombs don’t lose all their energy passing through the bulb: they merely lose most of their energy here due to the process of passing through a high resistance part of the circuit.

In part 3 of this series, we’ll look at how resistance can be modelled using the CTM.

You can read part 3 here.

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 Electric Circuits? Climb On Board The Coulomb Train!

I’ve said it before and I’ll say it again: teaching electric circuits is hard.

Providing your students with a conceptual model can, in my opinion, be immensely helpful. In recent years, I have used what I call the Coulomb Train Model (CTM). It is essentially a variation on the “stiff chain” analogies that some researchers have argued as being particularly useful in developing students’ understanding.

One reason why I like the CTM is that it provides a physical picture to aid students’ comprehension of many of the electrical equations needed at GCSE.

Of course, any analogy or model will have its flaws, but on the whole I think the CTM has fewer than many of its rivals!

Essentially, the CTM pictures an electric circuit as a continuously moving chain of postively-charged “trucks” called coulombs that carry energy from the cell to (say) the bulb. In the diagram below, they should be pictured as moving clockwise.

The coulomb is, of course, the S.I. unit of electric charge, so rest assured that there is method in the apparent madness of naming our “trucks” with a word that would be unfamiliar to most of our students.

Charge flow = current x time

Charge flow = number of coulombs that pass a given point in time.

Current = number of coulombs that pass by in one second (i.e. current = charge flow / time).

In other words, an ammeter counts the coulombs passing by in one second. The ammeter only “sees” the coulombs and does not register how much (or how little) energy each one contains. Therefore current I1 and current I2 are equal.

The ammeters are shown as being semi-transparent in order to provide a visual cue that they are low resistance devices.

Energy transferred = charge flow x potential difference

On the CTM, potential difference can be pictured as energy being added to, or removed from, each coulomb.

For example, if one joule is removed from each coulomb as they pass through the bulb, the potential difference across the bulb is one volt. If one joule is added to each coulomb as they pass through the cell, then the potential difference (or e.m.f.) across the cell is one volt.

The total energy transferred from (say) ten coulombs passing through the bulb would be charge flow (10 coulombs) x potential difference (1 volt) = 10 joules.

The white gloves on the voltmeter are intended to be reminiscent of the white gloves of a snooker referee.

The intention is to disrupt the flow of the coulombs as little as possible and so this is a visual reminder that a voltmeter is a high resistance instrument.

To emphasise the fact that potential difference is an “energy difference”, challenge students to predict the reading on this voltmeter.

The potential difference V3 is, of course, zero since there is no transfer of energy to or from the coulombs.

Current in Series and Parallel Circuits

I think the CTM can be really effective in allowing students to a see a comprehensible physical analogue of the circuits.

For example, I3 = I4 = I5 = 0.5 amps; I6 = I11 = 2 amps; and I7 = I8 = I9 = I10 = 1 amp.

Potential difference in series and parallel circuit

Equally, I think the CTM can give a comprehensible physical picture of the situation.

In this case (assuming the the cell has a p.d. of 1 V and the bulbs are identical), V4 = V5 = 0.5 V.

In the parallel circuit, each bulb tranfers one joule of energy from each bulb, and so the potential difference across each bulb is one volt.

You can read more on the CTM here.

Electrifying Engelmann

It is a long-standing and melancholy truth that, despite the best efforts of many legions of Physics teachers, many students continue to not only dislike electricity, but to hate it with the white-hot intensity of a million suns.

What we have here, I think, is a classic failure to communicate.

A final fact is that samenesses and differences of examples are more obvious when the examples are juxtaposed. This fact implies that the continuous conversion of examples provides the clearest presentation of samenesses and, differences because it creates the changes that occur from one example to the next.

— Siegfried Engelmann and Douglas Carmine, Theory of Instruction (1982) p.46

Looking at my own teaching, I certainly attempt to juxtapose a number of circuits. I really want to highlight the similarities and differences between circuits in order to better develop my students’ understanding. But the problem is that both limited resources and other practical considerations mean that the juxtapositioning cannot happen by continuous conversion, except very rarely.

For example, I would set up (or ask students to set up) a circuit with a single bulb with an ammeter, then I (or we) would disassemble the circuit and rebuild it with the ammeter in a different position, or a second bulb added in series or in parallel . . .

It occurs to me that what we are relying on to thread these juxtapositions together in students’ minds is a sequence of circuit diagrams. I suppose it’s another case of the curse of knowledge writ large: experts and novices think differently.

As a beginning teacher, I remember being genuinely shocked that many students found it easier to interpret a photograph or a 3D drawing rather than the nice, clutter-free, minimalist lines of a circuit diagram.

Without a doubt, many students retain strong visual impressions of many of the circuit diagrams they encounter, but they do not parse and decode the diagrams in the same way as their teachers do.

And that, I think, is the major problem when we are introducing electric circuits.

But what to do?

— R. S. Thomas, The Cure

Can we introduce the important aspects of electrical circuits by continuous conversion of examples?

I think we can. And what is more, I think it will be more effective than the itty-bitty assembly and disassembly of circuits that I have practiced to date.

Conservation of electrical current (and current in parallel circuits) by continuous conversion

Parallel Circuit

This is introduced with a teacher demonstration of the above circuit. Students are invited to note the identical readings on both ammeters and asked to explain why they are identical. They are then asked to predict the effect of adding a second bulb in parallel. The teacher then adds the second bulb by connecting the flying lead. The process is repeated with the third and fourth bulbs, with the teacher testing students’ understanding by asking them to predict the change in current readings as bulbs are added and removed. The teacher also tests students’ understanding of the conservation of current by asking students to predict whether the reading on both ammeters will be the same or different as bulbs are added and removed.

I find it useful to include a bulb that is not identical to the other three. It should be noticeably brighter or dimmer than the other three with the same p.d. so that students do not make the incorrect inference that the current always increases or decreases in equal steps when the circuit is changed.

The teacher could also draw the original circuit on a student whiteboard and ask students to do likewise. The changes that are about to be made could be described and students could be asked could alter the picture/circuit diagram and write their prediction on their whiteboards. They could then compare their version with the teacher’s and their prediction could be quickly tested by making the proposed changes “live” in front of the students.

If resources and time permit, students could then, of course, go on to construct their own parallel circuits as a class practical. However, I think it is important that these vital, foundational ideas are introduced (or re-introduced!) via a teacher demonstration to avoid possible cognitive overload for students.

Series circuits by continuous conversion

Series Circuit

In this demonstration circuit, four of the three bulbs are short-circuited so that they are initially unlit. The teacher asks students to explain only one bulb in the circuit is lit: it is helpful if they have previously encountered parallel circuits and can explain this in terms of electrical current taking the “easier” route (assuming they have not yet encountered the concept of electrical resistance).

Again, the two ammeters allow the teacher to emphasise and test students understanding of the idea that current is conserved.

The teacher then asks students to predict the change in current reading when switch X is opened: will it increase or decrease? Why would it increase or decrease? The process is repeated with switches Y and Z and students’ understanding is tested by asking them to predict the effect on the current reading of opening or closing X, Y or Z.

As before, the teacher would amend her circuit diagram on her student whiteboard and students would do likewise. For example: “I am going to open switch Y. Change the circuit diagram. Show me. What will happen to the reading on the left hand ammeter? What will happen to the reading on the right hand ammeter? Explain why.”

Again, I recommend that at least one out of the four bulbs in not identical to the other three to help prevent students from drawing the incorrect inference that the current will always increase or decrease in identical steps.