CTM

The Coulomb Train Model Revisited (Part 5)

e=mc2andallthat4 Comments

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.

The Coulomb Train Model Revisited (Part 4)

e=mc2andallthat3 Comments

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 epistemology. Science & Education, 24(5), 561-590.

The Coulomb Train Model Revisited (Part 3)

e=mc2andallthat5 Comments

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)

e=mc2andallthat4 Comments

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)

e=mc2andallthat8 Comments

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.

Potential Divider Circuits and the Coulomb Train Model

e=mc2andallthat12 Comments

A potential divider circuit is, essentially, a circuit where two or more components are arranged in series.

(a) Two resistors in series; (b) an ammeter (top) and an electric motor in series; (c) (L to R) a resistor, filament lamp and variable resistor in series

For non-physicists, these types of circuit can sometimes present problems, so in this post I am going to look in detail at the basic physics involved; and I am going to explain them using the CTM or Coulomb Train Model. (You can find the CTM model explained here.)

In the AQA GCSE Physics (and Combined Science) specifications, students are required to know that:

Extract from p.26 of the AQA spec

First, let’s look at the basics of describing electric circuits: current, potential difference and resistance.

1.0 Using the CTM to explain current, potential difference and resistance

Pupils tend to start with one concept for electricity in a direct current circuit: a concept labelled ‘current’, or ‘energy’ or ‘electricity’, all interchangeable and having the properties of movement, storability and consumption. Understanding an electrical circuit involves first differentiating the concepts of current, voltage and energy before relating them as a system, in which the energy transfer depends upon current, time and the potential difference of the battery.

The notion of current flowing in the circuit is one which pupils often meet in their introduction to a circuit and, because this relates well with their intuitive notions, this concept becomes the primary concept. (Driver 1994: 124 [italics added])

To my mind, the CTM is an excellent “bridging analogy” that helps students visualise the invisible. It is a stepping stone that provides some concrete representations of abstract quantities. In my opinion, it can help students

  1. move away from analysing circuits in terms of just current. (In my experience, even when students use terms like “potential difference”, in their eyes what they call “potential difference” behaves in a remarkably similar way to current e.g. it “flows through” components.)
  2. understand the difference between current, potential difference and resistance and how important each one is
  3. begin thinking of a circuit as a whole, interconnected system.

1.1 The CTM and electric current

Let’s begin by looking at a very simple circuit: a one ohm resistor connected across a 1 V cell.

A simple circuit
A very simple circuit

Note that it is a good teaching technique to include two ammeters on either side of the component, although the readings on both will be identical. This is to challenge the perennial misconception that electric current is “used up”. Electric charge, according to our current understanding of the universe, is a conserved quantity like energy in that it cannot be created or destroyed.

The Coulomb Train Model invites us to picture an electric circuit as a flow of positively charged coulombs carrying energy around the circuit in a clockwise fashion as shown below. The coulombs are linked together to form a continuous chain.

The CTM version of a simple circuit
The CTM applied to the very simple circuit shown above.

The name coulomb is not chosen at random: it is the SI unit of electric charge.

The current in this circuit will be given by I = V / R (equation 18 in the list on p.96 of the AQA spec, if you’re keeping track).

Using the AQA mark scheme-friendly FIFA protocol:

The otherwise inexplicable use of the letter “I” to represent electric current springs from the work André-Marie Ampère (1775–1836) and the French phrase intensité de courant (intensity of current).

From Q = I t (equation 17, p.96), current is a flow of electric charge, since I = Q / t. That is to say, if a charge of 2 coulombs passes (AQA call this a “charge flow”) in 2 seconds, the current will be …

A current of 1 amp is therefore represented on the CTM as 1 coulomb (or truck) passing by each second.

1.2 The CTM and Potential Difference

Potential difference or voltage is essentially the “energy difference” across any two parts of a circuit.

The equation used to define potential difference is not the familiar V = IR but rather the less familiar E = QV (equation 22 in the AQA list) where E is the energy transferred, Q is the charge flow (or the number of coulombs passing by in a certain time) and t is the time in seconds.

Let’s see what this would look like using the CTM:

(a) Circuit diagram showing how the measure the potential difference across a 1 V cell. (b) The same circuit represented using the CTM. (Note that the “white gloves” on the ends of the voltmeter connections are intended to be reminiscent of the white gloves of a snooker referee, indicating that the voltmeter does not disrupt the flow of the coulombs: in other words, the voltmeter has a high resistance.)

For the circuit shown, the voltmeter reading is 1 volt.

Note that on the CTM representation, one joule of energy is added to each coulomb as it passes through the cell.

If we had a 1.5 V cell then 1.5 joules would be transferred to each coulomb as it passed through, and so on.

(a) Circuit diagram showing potential difference measured across a connector with negligible resistance. (b) The same circuit represented using the CTM

If the voltmeter is moved to a different position as shown above, then the reading is 0 volts. This is because the coulombs at the points “sampled” by the voltmeter have the same amount of energy, so there is zero energy difference between them.

(a) Measuring the potential difference across a resistor. (b) The same circuit shown using the CTM.

In the position shown above, the voltmeter is measuring the potential difference across the resistor. For the circuit shown (assuming negligible resistance in all other parts of the circuit) the potential difference will be 1 V. In other words, each coulomb is losing one joule of energy as it passes through the resistance.

1.3 The CTM and Resistance

(a) Measuring the current through and the potential difference across a resistor. (b) The same circuit represented using the CTM.

In the circuit above, the potential difference across the resistor is 1 V and the current is 1 amp.

Resistance can therefore be thought of as the potential difference required to drive a current of 1 amp through that part of the circuit. It can also be thought of as the energy lost by each coulomb when a current of 1 amp flows through that part of the circuit; or, energy lost per coulomb per amp.

1.4 Summary

On the diagrams below, the coulombs are moving clockwise.

Summary of CTM

2.0 The CTM applied to a potential divider circuit

A potential divider circuit simply means that at least two resistors are in series so that the potential difference of the cell is shared across the resistors.

2.1 Two identical resistors

Because the two resistors are identical, the 3 V supply is shared equally across both resistors. That is to say, there is a potential difference of 1.5 V across each resistor. But let’s check this by applying V = IR (eq. 18). The total potential difference is 3 V and the total resistance is 1 ohm + 1 ohm = 2 ohms.

Now let’s use V = IR to check that the potential difference across each separate resistor is indeed half the total supply of 3 V. The resistance of one resistor is one ohm and the current through each one is 1.5 A. So V = 1.5 x 1 = 1.5 V.

But what would happen if we doubled the value of each resistor to 2 ohms?

Well, the current would be smaller: I = V/R = 3/4 = 0.75 amps.

The potential difference across each separate resistor would be V = I R = 0.75 x 2 = 1.5 V

So, the potential difference is always split equally when two identical resistors are placed in series (although, of course, the total resistance and the current will be different depending on the values of the resistors).

2.2a Two non-identical resistors

Let’s consider a circuit with a 2 ohm resistor in series with a 1 ohm resistor.

In this circuit, the total resistance is 1 ohm + 2 ohms = 3 ohms. The current flowing through the circuit is I = V / R = 3 / 3 = 1 amp.

So the potential difference across the 2 ohm resistor is V = IR = 1 x 2 = 2 V and the potential difference across the one ohm resistor is V = IR = 1 x 1 = 1 V.

Note that the resistor with the largest value gets the largest “share” of the potential difference.

2.2b Two non-identical resistors (different order)

Now let’s reverse the order of the resistors.

The current remains unchanged because the total resistance of the circuit is still the same.

Note that the largest resistor still gets the largest share of the potential difference, whichever way round the resistors are placed.

2.3 In Defence of the CTM and Donation Models

Many Physics teachers prefer “rope models” to so-called “donation models” like the CTM.

And it is perfectly true that rope models have some good points such as the ability to easily explain AC and a more accurate approximation of what happens when current starts to flow or stops flowing. The difficulty in their use, in my opinion, is that you are using concepts that many students barely understand (e.g. friction to model resistance) to explain how very unfamiliar concepts such as potential difference work. Also, the vagueness of some of the analogs is unhelpful: for example, when we compare potential difference to “push”, are we talking about the net resultant force on the rope or simply the force needed to balance the frictional force and keep it moving at a steady speed?

To my way of thinking, the CTM has the advantage of encouraging quantitative thinking about current, potential difference and resistance almost from the moment of first teaching. Admittedly, it cannot cope with AC — but then again, we model AC as a direct current when we use RMS values. Now admittedly, rope models are far better at picturing what happens in the initial fractions of a second when a current starts to flow after closing a switch. Be that as it may, the CTM comes into its own when we consider the “steady state” of current flow after the initial surge currents.

One of the frequent criticisms (which is usually considered quite damning) of this type of model is “How do the coulombs know how much energy to drop off at each resistor?”

For example, in the diagram above, how do the coulombs “know” to drop off 1 J at the first resistor and 2 J at the second resistor?

The answer is: they don’t. Rather, the energy loss is due to the nature of the resistor: think of a resistor as a tunnel lined with strip curtains. A coulomb loses only a small amount of its excess energy passing through a low value resistor, but a much larger amount passing through a higher value resistor, as modelled below.

Strip curtain model for CTM
A 1 ohm and 2 ohm resistor modelled as strip curtains

FWIW I therefore commend the use of the CTM to all interested parties. 

References

Driver, R., Squires, A., Rushworth, P., & Wood-Robinson, V. (1994). Making sense of secondary science: Research into children’s ideas. Routledge.

Teaching Electric Circuits? Climb On Board The Coulomb Train!

e=mc2andallthat21 Comments

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.

You can also a find a booklet to support teaching electric circuits using the CTM here.