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‘.