Explaining current flow in conductors (part two)

Do we delve deeply enough into the actual physical mechanism of current flow through electrical conductors using the concepts of charge carriers and electric fields in our treatments for GCSE and A-level Physics? I must reluctantly admit that I am increasingly of the opinion that the answer is no.

In part one we discussed two common misconceptions about the physical mechanism of current flow, namely:

  1. The all-the-electrons-in-a-conductor-repel-each-other misconception; and
  2. The electric-field-of-the-battery-makes-all-the-charge-carriers-in-the-circuit-move misconception.

What, then, does produce the internal electric field that drives charge carriers through a conductor?

Let’s begin by looking at the properties that such a field should have.

Current and electric field in an ohmic conductor

(You can see a more rigorous derivation of this result in Duffin 1980: 161.)

We can see that if we consider an ohmic conductor then for a current flow of uniform current density J we need a uniform electric field E acting in the same direction as J.

What produces the electric field inside a current-carrying conductor?

The electric field that drives charge carriers through a conductor is produced by a gradient of surface charge on the outside of the conductor.

Rings of equal charge density (and the same sign) contribute zero electric field at a location midway between the two rings, whereas rings of unequal charge density (or different sign) contribute a non-zero field at that location.

Sherwood and Chabay (1999): 9

These rings of surface charge produce not only an internal field Enet as shown, but also external fields than can, under the right circumstances, be detected.

Relationship between surface charge densities and the internal electric field

Picture a large capacity parallel plate capacitor discharging through a length of high resistance wire of uniform cross section so that the capacitor takes a long time to discharge. We will consider a significant period of time (a small fraction of RC) when the circuit is in a quasi-steady state with a current density of constant magnitude J. Since E = J / σ then the internal electric field Enet produced by the rings of surface charge must be as shown below.

Schematic diagram showing the relationship between the surface charge density and the internal electric field

In essence, the electric field of the battery polarises the conducting material of the circuit producing a non-uniform arrangement of surface charges. The pattern of surface charges produces an electric field of constant magnitude Enet which drives a current density of constant magnitude J through the circuit.

As Duffin (1980: 167) puts it:

Granted that the currents flowing in wires containing no electromotances [EMFs] are produced by electric fields due to charges, how is it that such a field can follow the tortuous meanderings of typical networks? […] Figure 6.19 shows diagrammatically (1) how a charge density which decreases slowly along the surface of a wire produces an internal E-field along the wire and (2) how a slight excess charge on one side can bend the field into the new direction. Rosser (1970) has shown that no more than an odd electron is needed to bend E around a ninety degree corner in a typical wire.

Rosser suggests that for a current of one amp flowing in a copper wire of cross sectional area of one square millimetre the required charge distribution for a 90 degree turn is 6 x 10-3 positive ions per cm3 which they call a “minute charge distribution”.

Observing the internal and external electric fields of a current carrying conductor

Jefimenko (1962) commented that at the time

no generally known methods for demonstrating the structure of the electric field of the current-carrying conductors appear to exist, and the diagrams of these fields can usually be found only in the highly specialized literature. This […] frequently causes the student to remain virtually ignorant of the structure and properties of the electric field inside and, especially, outside the current-carrying conductors of even the simplest geometry.

Jefimenko developed a technique involving transparent conductive ink on glass plates and grass seeds (similar to the classic linear Nuffield A-level Physics electrostatic practical!) to show the internal and external electric field lines associated with current-carrying conductors. Dry grass seeds “line up” with electric field lines in a manner analogous to iron filings and magnetic field lines.

Photograph from Jefimenko (1962: 20). Annotations added

Coming soon . . .

In part 3, we will analyse the transient processes by which these surface charge distributions are set up.

References

Duffin, W. J. (1980). Electricity and magnetism (3rd ed.). McGraw Hill Book Co.

Jefimenko, O. (1962). Demonstration of the electric fields of current-carrying conductorsAmerican Journal of Physics30(1), 19-21.

Rosser, W. G. V. (1970). Magnitudes of surface charge distributions associated with electric current flow. American Journal of Physics38(2), 265-266.

Sherwood, B. A., & Chabay, R. W. (1999). A unified treatment of electrostatics and circuits. URL http://cil. andrew. cmu. edu/emi. (Note: this article is dated as 2009 on Google Scholar but the text is internally dated as 1999)

Explaining current flow in conductors (part one)

Do we delve deeply enough into the actual physical mechanism of current flow through electrical conductors (in terms of charge carriers and electric fields) in our treatments for GCSE and A-level Physics? I must reluctantly admit that I am increasingly of the opinion that the answer is no.

Of course, as physics teachers we talk with seeming confidence of current, potential difference and resistance but — when push comes to shove — can we (say) explain why a bulb lights up almost instantaneously when a switch several kilometres away is closed when the charge carriers can be shown to be move at a speed comparable to that of a sedate jogger? This would imply a time delay of some tens of minutes between closing the switch and energy being transferred from the power source (via the charge carriers) to the bulb.

When students asked me about this, I tended to suggest one of the following:

  • “The electrons in the wire are repelling each other so when one close to the power source moves, then they all move”; or
  • “Energy is being transferred to each charge carrier via the electric field from the power source.”

However, to be brutally honest, I think such explanations are too tentative and “hand wavy” to be satisfactory. And I also dislike being that well-meaning but unintentionally oh-so-condescending physics teacher who puts a stop to interesting discussions with a twinkly-eyed “Oh you’ll understand that when you study physics at degree level.” (Confession: yes, I have been that teacher too often for comfort. Mea culpa.)

Sherwood and Chabay (1999) argue that an approach to circuit analysis in terms of a predominately classical model of electrostatic charges interacting with electric fields is very helpful:

Students’ tendency to reason locally and sequentially about electric circuits is directly addressed in this new approach. One analyzes dynamically the behaviour of the *whole* circuit, and there is a concrete physical mechanism for how different parts of the circuit interact globally with each other, including the way in which a downstream resistor can affect conditions upstream.

(Side note: I think the Coulomb Train Model — although highly simplified and applicable only to a limited set of “steady state” situations — is consistent with Sherwood and Chabay’s approach, but more on that later.)

Misconception 1: “The electrons in a conductor push each other forwards.”

On this model, the flowing electrons push each other forwards like water molecules pushing neighbouring water molecules through a hose. Each negatively charged electron repels every other negatively charged electron so if one free electron within the conductor moves, then the neighbouring free electrons will also move. Hence, by a chain reaction of mutual repulsion, all the electrons within the conductor will move in lockstep more or less simultaneously.

The problem with this model is that it ignores the presence of the positively charged ions within the metallic conductor. A conveniently arranged chorus-line of isolated electrons would, perhaps, behave analogously to the neighbouring water molecules in a hose pipe. However, as Sherwood and Chabay argue:

Averaged over a few atomic diameters, the interior of the metal is everywhere neutral, and on average the repulsion between flowing electrons is canceled by attraction to positive atomic cores. This is one of the reasons why an analogy between electric current and the flow of water can be misleading.

The flowing electrons inside a wire cannot push each other through the wire, because on average the repulsion by any electron is canceled by the attraction of a nearby positive atomic core (Diagram from Sherwood and Chabay 1999: 4)

Misconception 2: “The charge carriers move because of the electric field from the battery.”

Let’s model the battery as a high-capacity parallel plate capacitor. This will avoid the complexities of having to consider chemical interactions within the cells. Think of a “quasi-steady state” where the current drawn from the capacitor is small so that electric charge on the plates remains approximately constant; alternatively, think of a mechanical charge transfer mechanism similar to the conveyor belt in a Van de Graaff generator which would be able to keep the charge on each plate constant and hence the potential difference across the plates constant (see Sherwood and Chabay 1999: 5).

A representation of the electric field around a single cell battery (modelled as a parallel plate capacitor)

This is not consistent with what we observe. For example, if the charge-carriers-move-due-to-electric-field-of the-battery model was correct then we would expect a bulb closer to the battery to be brighter than a more distant bulb; this would happen because the bulb closer to the battery would be subject to a stronger electric field and so we would expect a larger current.

A bulb closer to the battery is NOT brighter than a bulb further away from the battery (assuming negligible resistance in the connecting wires)

There is the additional argument if we orient the bulb so that the current flow is perpendicular to the electric field line, then there should be no current flow. Instead, we find that the orientation of the bulb relative to the electric field of the battery has zero effect on the brightness of the bulb.

There is no change of brightness as the orientation of the bulb is changed with respect to the electric field lines from the battery

Since we do not observe these effects, we can conclude that the electric field lines from the battery are not solely responsible for the current flow in the circuit.

Understanding the cause of current flow

If the electric field of the battery is not responsible on its own for the potential difference that causes a current to flow, where does the electric field come from?

Interviews reveal that students find the concept of voltage difficult or incomprehensible. It is not known how many students lose interest in physics because they fail to understand basic concepts. This number may be quite high. It is therefore astonishing that this unsatisfactory situation is accepted by most physics teachers and authors of textbooks since an alternative explanation has been known for well over one hundred years. The solution […] was in principle discovered over 150 years ago. In 1852 Wilhelm Weber pointed out that although a current-carrying conductor is overall neutral, it carries different densities of charges on its surface. Recognizing that a potential difference between two points along an electric circuit is related to a difference in surface charges [is the answer].

Härtel (2021): 21

We’ll look at these interesting ideas in part two.

[Note: this post edited 10/7/22 because of a rewritten part two]

References

Härtel, H. (2021). Voltage and Surface ChargesEuropean Journal of Physics Education12(3), 19-31.

Sherwood, B. A., & Chabay, R. W. (1999). A unified treatment of electrostatics and circuits. URL http://cil. andrew. cmu. edu/emi. (Note: this article is dated as 2009 on Google Scholar but the text is internally dated as 1999)

Acknowledgements

The circuit representations were produced using the excellent PhET Sims circuit simulator.

I was “awoken from my dogmatic slumbers” on this topic (and alerted to Sherwood and Chabay’s treatment) by Youtuber Veritasium‘s provocative videos (see here and here).

Modelling Electrical Power Equations using the CTM

The Coulomb Train Model (CTM) is a straightforward, easily pictured representation that helps novice learners develop an initial “sense of mechanism” about how electric circuits work. You can read about it here and here (and, to some extent, track its development over time).

In this post, however, I want to focus on how effective the CTM is in helping students understand the energy and power formulas associated with electric circuits: notably E = QV, P = IV and P=I2R.

E=QV and the CTM

An animated version of the Coulomb Train Model

The E in E=QV stands for the energy transferred to the bulb by the electric current (in joules, J). The Q is the charge flow in coulombs, C. The V is the potential difference across the resistor in volts, V.

Using the Coulomb Train Model:

  • Each grey truck passing through the bulb represents one coulomb of charge flow. Q is therefore the number of grey trucks passing through the bulb in a certain time t. (We won’t specify what that time t is now but we will return to it shortly.)
  • The potential difference V is the energy transferred out of each coulomb as they pass through the bulb. If one joule (represented by the orange stuff in the truck) is transferred from each coulomb then the potential difference is one volt. If two joules then the potential difference is two volts, and so on.

How can we increase the energy transferred into the bulb? There are two ways:

  1. Increase the total number of coulombs passing through the bulb. That is to say, increasing Q. We could do this by (a) waiting a longer time so that more coulombs pass through the bulb; or (b) increasing the current so that more coulombs pass through each second.
  2. Increase the energy transferred from each coulomb into the bulb. That is to say, increasing V. We could do this by increasing the potential difference of the cell so that each coulomb is loaded up with more energy.
Hewitt representation of changing the values of Q and V while keeping the other fixed

Or, of course, we could increase the values of Q and V simultaneously.

Hewitt representation of increasing Q and V simultaneously

All you need is E = Q V

In other words, the energy transferred per second (or the power P in watts, W) is equal to the product of the current I in amperes (or coulombs per second) and the potential difference V in volts, V.

A higher current will increase the power transferred to the bulb: more coulombs will pass through the bulb per second so more energy is transferred to the bulb each second. This can be modelled using the Coulomb Train Model as shown:

Using the CTM to show the relationship between current I and power P

Increasing the potential difference V (i.e. the energy carried by each coulomb) would also increase P.

Deriving P=I2R from P=IV

If we start with P=IV but remember that V=IR then P=I(IR) so P=I2R.

This can be represented on the Coulomb Train Model like this:

We can increase the power transferred to the resistor by:

  • Increasing the value of the resistor (and keeping I constant, which implies that V would have to be increased). Doubling the value of R would double the value of P.
  • Increasing the value of I. However, since the formula includes I squared then this would have a disproportionate effect on P. For example, if I was doubled then P would be quadrupled. A Hewitt representation can be useful for highlighting this to students; for example:
A Hewitt representation of the effect of doubling I on P

Linking the electrical power formulas

The electrical power equations when considered in isolation can seem random and unconnected. Making the links between them explicit can be not just a powerful aid to memory, but also hints at the power and coherence of that noble exploration of reality and possibility called physics.

Series and Parallel Circuits — an unhelpful dichotomy?

Anakin Skywalker and Obi Wan Kenobi discuss the possible unhelpfulness of the concept of ‘series circuits’ and ‘parallel circuits

Are physics teachers following the Way of the Sith? Are we all crossing over to the Dark Side when we talk about ‘series circuits’ and ‘parallel circuits’?

I think that, without meaning to, we may be presenting students with what amounts to a false dichotomy: that all circuits are either series circuits or parallel circuits.

Venn diagram showing the false dichotomy view of series and parallel circuits

The actual situation is more like this:

A Venn diagram showing a more nuanced and realistic view of series and parallel circuits

The confusion may stem from our usage of the word ‘circuit’: are we referring holistically to the entire assemblage of components (highlighted in red) or the individual ‘complete circuits’ (highlighted in green and blue)?

Will the actual ‘circuit’ please stand up? The red circuit is a hybrid circuit, the green circuit is a series circuit, and the blue circuit shows a single resistor in series or parallel with cell (depending on how you look at it)

How to avoid the false dichotomy

I think we should always refer to components in series or components in parallel rather than ‘series circuits’ or ‘parallel circuits’.

Teaching components in parallel using the ‘all-in-a-row’ circuit convention

I’ve written before about what I think is the confusing ‘hidden rotation’ present in normal circuit diagrams. I find redrawing circuit diagrams using the ‘all-in-a-row’ convention useful for explaining circuit behaviour. For simplicity, we’ll assume that all the resistors in the diagrams that follow have a resistance of one ohm.

This can be shown using the Coulomb Train Model like this (coulombs pictured as moving clockwise):

The reason the voltmeter across the cell reads +1.5 V is that energy is being transferred from the chemical energy store of the cell *into* the coulombs. The reason the voltmeter reads -1.5 V across the resistor is that energy is being transferred *from* the coulombs and into the thermal energy store of the resistor.

The current passing through the resistor using I = V/R = 1.5 V / 1 = 1.5 amperes.

Now let’s apply this convention when two resistors are in parallel.

This can be represented using the Coulomb Train Model like this:

I think it’s far clearer that ammeter W is measuring the total current in the circuit while X and Y are measuring the ‘part-current’ passing through R1 and R2 using this convention. (Note: we are assuming that each resistor has a resistance of one ohm.)

Each resistor has a potential difference of -1.5 V because 1.5 J of energy is being shifted from each coulomb as they pass through each resistor.

Also, it is clearer that the cell’s chemical energy store is being drained more quickly when there are two resistors in parallel: two coulombs have to be filled with 1.5 J of energy for each one coulomb in the single resistor circuit.

Thinking about current, the total current in the circuit is 3.0 amperes; so the resistance R = V / I = 1.5 / 3.0 = 0.5 ohms. So two resistors in parallel have a smaller resistance than a single resistor — this is a result that is well worth emphasising for students as so many of them find this completely counterintuitive!

Teaching components in series using the all-in-a-row convention

This circuit can be represented using the Coulomb Train Model like this:

The pattern of potential difference can be explained by looking at the orange ‘energy levels’ carried by each coulomb.

A current of one amp is one coulomb passing per second, so we can see that an ammeter reading would have the same value wherever the ammeter is placed in the circuit.

But look closely at R1: it only has 0.75 V of potential difference across. From I = V/R = 0.75 / 1 = 0.75 amperes.

This means that the total resistance of the circuit from R = V/I is, of course, 2 ohms.

Conclusion

I regret to say that I have probably been teaching ‘series circuits’ and ‘parallel circuits’ on autopilot for much of my career; the same may even be true of some readers of this blog(!)

The Coulomb Train Model has been considered in depth in previous blogs, but I think it’s a good model to encourage students to use their physical intuition (aka ’embodied cognition’) to understand electric circuits.

Whether you agree with the suggested outlines above or not, I hope that it has given you some fruitful food for thought.

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!

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 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.