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).

Speed of sound using Phyphox

You can get encouragingly accurate values for the speed of sound in the school laboratory using a tape measure and two smartphones (or tablets) running the Phyphox app.

Phyphox (pronounced FEE-fox) is an award-winning free app that was developed by physicists at Aachen University who wanted to give users direct access to the many sensors (e.g. accelerometers and magnetometers) which are standard features on many smartphones. In effect, it turns even the humblest smartphone or tablet into a multifunctional measuring instrument comparable to one of Star Trek’s famous ‘tricorders’.

Star Trek Tricorder (not Phyphox)

To measure the speed of sound, we need two smartphones or tablets running Phyphox. We will be using the ‘Acoustic Stopwatch’ which measures the time between two acoustic events.

Phyphox main screen

Step 1: Place two devices a measured distance s apart. Typically about 2 or 3 m should be OK otherwise the sound made by A will not be loud enough to control stopwatch B — this can be established through trial and error and depends on many factors including the background noise level.

Step 2: Person A makes a loud sound (a clap or a single syllable shout like ‘Hey!’ is good).

Step 3: Person B and stopwatch B wait for sound created by A to reach them.
Step 4: The sound reaches stopwatch B and starts it running and B hears the sound.

Step 5: B makes a loud sound in response.

Step 6: The loud sound made by B reaches stopwatch B and makes it stop. Let’s call the time displayed tB. This measures the delay between the sound from A reaching stopwatch B and B reacting to the sound and stopping the clock. It includes the time taken for the initial sound travelling from the device to B, B’s reaction time, and the time taken for the sound made by B to travel to stopwatch B. B does not have to be particularly ‘quick off the mark’ to respond to A’s sound — although the shorter the time then the less likely it is then a background noise will interrupt the experiment.

Step 7: The sound made by B travels toward stopwatch A.

Step 8: The sound made by B reaches stopwatch A and makes it stop. Let’s call the time recorded on stopwatch A tA.

If we break down the events included in tA and tB, we find that tA is always larger than tB:

If we subtract tAtB we find that this is the time it takes sound to travel a distance of 2s.

Step 9: We can therefore use this formula to find the v the speed of sound.

We have found that this method works well giving mean values of about 350 m/s for the speed of sound (which will vary with air temperature). This video models the method.

And so we have a reasonably practicable method of measuring the speed of that doesn’t involve complex equipment that is unfamiliar to most students; or a method that involves finding a large and featureless wall that produces a detectable echo when a loud sound is made from a point several metres in front of it.

I don’t know about you, but as a physics teacher, I feel cheated. If it doesn’t involve a double beam oscilloscope, a signal generator, two microphones and two power amplifiers then I simply don’t want to know about it . . .

The Rite of AshkEnte, quite simply, summons and binds Death.  Students of the occult will be aware that it can be performed with a simple incantation, three small bits of wood and 4cc of mouse blood, but no wizard worth his pointy hat would dream of doing anything so unimpressive; they knew in their hearts that if a spell didn’t involve big yellow candles, lots of rare incense, circles drawn on the floor with eight different colours of chalk and a few cauldrons around the place then it simply wasn’t worth contemplating.

Terry Pratchett, ‘Mort’ (1987)

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.

Through a glass, lightly

Any sufficiently advanced technology is indistinguishable from magic.

Profiles of the Future (1962)

So wrote Arthur C. Clarke, science fiction author and the man who invented the geosynchronous communications satellite.

Clarke later joked of his regret at the billions of dollars he had lost by not patenting the idea, but one gets the impression that what really rankled him was another (and, to be fair, uncharacteristic) failure of imagination: he did not foresee how small and powerful solid state electronics would become. He had pictured swarms of astronauts crewing vast orbital structures, having their work cut out as they strove to maintain and replace the thousands of thermionic valves burning out under the weight of the radio traffic from Earth . . .

A communications nexus that could fit into a volume the size of a minibus, then (as the technology developed) a suitcase and then a pocket seemed implausible to him — and to pretty much everyone else as well.

And yet, here we are, living in the world that microelectronics has wrought. We are truly in an age of Clarkeian magic: our technology has become so powerful, so reliable and so ubiquitous — and so few of us have a full understanding of how it actually works — that it is very nearly indistinguishable from magic.

I want to outline a fictional vignette on the same theme which Clarke wrote in 1961 which has stayed with me since I read it. Please bear with me while I sketch out the story — it will lend some perspective to what follows.

In the novel A Fall of Moondust (1961) set in the year 2100, the lunar surface transport Selene has become trapped under fifteen metres of moondust. Rescue teams on the surface have managed to drill down to the stricken vessel with a metal pipe to supply the unfortunate passengers and crew with oxygen from the rescue ship. Communication would seem to be impossible as the Selene’s radio has been destroyed, but luckily Chief Engineer (Earthside) Lawrence has a plan:

They would hear his probe, but there was no way in which they could communicate with him. But of course there was. The easiest and most primitive means of all, which could be so readily overlooked after a century and a half of electronics.

A few hours later, the rescue team’s pipe drills through the roof of the sunken vessel.

The brief rush of air gave everyone a moment of unnecessary panic as the pressure equalised. Then the pipe was open to the upper world, and twenty two anxious men and women waited for the first breath of oxygen to come gushing down it.

Instead, the tube spoke.

Out of the open orifice came a voice, hollow and sepulchral, but perfectly clear. It was so loud, and so utterly unexpected, that a gasp of surprise came from the company. Probably not more than half a dozen of these men and women had ever heard of a ‘speaking tube’; they had grown up in the belief that only through electronics could the voice be sent across space. This antique revival was as much a novelty to them as a telephone would have been to an ancient Greek.

The humble converging lens as an ‘antique revival’

If you hold a magnifying glass (or any converging lens) in front of a white screen, then it will produce a real, inverted image of any bright objects in front of it. This simple act can, believe it or not, draw gasps of surprise from groups of our ‘digital native’ students: they assume that images can only be captured electronically. The fact that a shaped piece of glass can do so is as much a novelty to them as an LCD screen would have been to Galileo.

Think about it for a moment: how often has one of our students seen an image projected by a lens onto a passive screen? The answer is: possibly never.

The cinema? Not necessarily — many cinemas use large electronic screens now; there is no projection room, no projector painting the action on the screen from behind us with ghostly, dancing fingers of light. School? In the past, we had overhead projectors and even interactive whiteboards had lens systems, but these have largely been replaced by LED and LCD screens.

I believe that if you do not take the time to show the phenomenon of a single converging lens projecting a real image on to a passive white screen to your students, they are likely to have no familiar point of reference on which to build their understanding and lens diagrams will remain a puzzling set of lines that have little or no connection to their world.

Teaching Ray Diagrams

Start with a slide that looks something like this:

What represents the lens? The answer is not the blue oval. On ray diagrams, the lens is represented by the vertical dotted line. F1 and F2 are the focal points of this converging lens and they are each a distance f from the centre of the lens, where f is the focal length.

Now what happens to a light ray from the object that passes through the optical centre of the lens?

The answer, of course, is a big fat nothing. Light rays which pass through the optical centre of a thin lens are undeviated.

Now let’s track what happens to a light ray that travels parallel to the principal axis as shown?

Make sure that your students are aware that this light ray hasn’t ‘missed’ the lens. The lens is the vertical dotted line, not the blue oval. What will happen is that it will be deviated so that it passes through F1 (this is because this is a converging lens; if it had been a diverging lens then it would be be bent so that it appeared to come from F2).

The image is formed where the two light rays cross, as shown below.

We can see that the image is inverted and reduced.

The image is formed close to F1 but not precisely at F1. This is because, although the object is distant from the lens (‘distant’ in this case being ‘further than 2f away’) it is not infinitely far away. However, the further we move the object away from the lens, then the closer to F1 the image is formed. The image will be formed a distance f from the screen when the object in very, very, very large distance away — or an ‘infinite distance’ away, if you prefer.

One of my physics teachers liked to say that ‘Infinity starts at the window sill’. In the context of thinking about lenses, I think he was right . . .

Method for finding the focal length of a converging lens (image from https://www.youtube.com/watch?v=AElLVGW9kxQ)

Some free stuff . . .

The PowerPoint that I used to produce the ray diagrams above is here. It is imperfect in a lot of ways but. truth be told, it has served me well over a number of years. It also features some other slides and animations that you may find useful — enjoy!

Postscript: ‘Through a glass, darkly’

The phrase ‘Through a glass, darkly’ comes from the writings of the apostle Paul:

For now we see through a glass, darkly; but then face to face: now I know in part; but then shall I know even as also I am known.

KJV 1 Corinthians 13:12

The New International Version translates the phrase less poetically as ‘Now we see but a poor reflection as in a mirror.’

It has been argued that the ‘glass’ Paul was referring to were pieces of naturally-occurring semi-transparent mineral that were used in the ancient world as lenses or windows. They tended to produce a recognisable but distorted view of the world — hence, ‘darkly’.

Better technology means that there is much less distortion produced by our glasses — hence, ‘through a glass, lightly’.

Visualising How Transformers Work

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

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

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

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

The basic principle of a transformer

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

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

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

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

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

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

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

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

Stepping Down

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

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

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

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

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

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

Stepping up

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

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

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

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

Last nail in the coffin of the formula triangle…

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

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

The Coulomb Train Model Revisited (Part 5)

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

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

Without further ado, here is a a summary.


A circuit with one resistor

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

This can be represented on the CTM like this:

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

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

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


A circuit with two resistors in series

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

This can be shown using the CTM like this:

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

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

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

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

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


Looking at the resistors individually

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

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

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

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

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

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

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

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

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


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

Simple answer: they don’t.

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

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

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

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

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

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

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


Afterword

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

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

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.

FIFA and Really Challenging GCSE Physics Calculations

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

FIFA stands for:

  • Formula
  • Insert values
  • Fine-tune
  • Answer

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


Framing FIFA as a modular approach

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

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

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


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

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

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

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

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

Edward Gibbon, The Decline and Fall of the Roman Empire

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


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

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

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

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

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

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

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

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


Conclusion: FIFA fixes it

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

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

Update: Ed Southall makes a persuasive case against formula triangles in this 2016 article.

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.