Monthly Archives: August 2019

Potential Divider Circuits and the Coulomb Train Model

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

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

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

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

Extract from p.26 of the AQA spec

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

 

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

 

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

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


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

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


1.1 The CTM and electric current

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

A simple circuit

A very simple circuit

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

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

The CTM version of a simple circuit

The CTM applied to the very simple circuit shown above.

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

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

Using the AQA mark scheme-friendly FIFA protocol:

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

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

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

1.2 The CTM and Potential Difference

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

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

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

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

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

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

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

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


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

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

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

1.3 The CTM and Resistance

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

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

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

1.4 Summary

On the diagrams below, the coulombs are moving clockwise.

Summary of CTM

2.0 The CTM applied to a potential divider circuit

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

2.1 Two identical resistors

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

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

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

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

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

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

2.2a Two non-identical resistors

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

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

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

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

2.2b Two non-identical resistors (different order)

Now let’s reverse the order of the resistors.

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

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

2.3 In Defence of the CTM and Donation Models

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

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

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

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

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

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

Strip curtain model for CTM

A 1 ohm and 2 ohm resistor modelled as strip curtains

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



References

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

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The Acceleration Required Practical Without Light Gates (And Without Tears)

Introduction

The AQA GCSE Required Practical on Acceleration (see pp. 21-22 and pp. 55-57) has proved to be problematic for many teachers, especially those who do not have access to a working set of light gates and data logging equipment.

In version 3.8 of the Practical Handbook (pre-March 2018), AQA advised using the following equipment featuring a linear air track (LAT). The “vacuum cleaner set to blow”, (or more likely, a specialised LAT blower), creates a cushion of air that minimises friction between the glider and track.

Screenshot 2019-08-08 at 14.47.50.png

However, in version 5.0 (dated March 2018) of the handbook, AQA put forward a very different method where schools were advised to video the motion of the car using a smartphone in an effort to obtain precise timings at the 20 cm, 40 cm and other marks.

Screenshot 2019-08-08 at 13.54.07.png

It is possible that AQA published the revised version in response to a number of schools contacting them to say. “We don’t have a linear air track. Or light gates. Or a ‘vacuum cleaner set to blow’.”

The weakness of the “new” version (at least in my opinion) is that it is not quantitative: the method suggested merely records the times at which the toy car passed the lines. Many students may well be able to indirectly deduce the relationship between resultant force and acceleration from this raw timing data; but, to my mind, it would be cognitively less demanding if they were able to compare measurements of resultant force and acceleration instead.

Adapting the AQA method to make it quantitative

Screenshot 2019-08-08 at 15.39.40.png

We simplify the AQA method as above: we simply time how long the toy car takes to complete the whole journey from start to finish.

If a runway of one metre or longer is set up, then the total time for the journey of the toy car will be 20 seconds or so for the smallest accelerating weight: this makes manual timing perfectly feasible.

Important note: the length of the runway will be limited by the height of the bench. As soon as the weight stack hits the floor, the toy car will no longer experience an accelerating force and, while it may continue at a constant speed (more or less!) it will no longer be accelerating. In practice, the best way to sort this out is to pull the toy car back so that the weight stack is just below the pulley and mark this position as the start line; then slowly move the toy car forward until the weight stack is just touching the floor, and mark this position as the finish line. Measure the distance between the two lines and this is the length of your runway.

In addition, the weight stack should feature very small masses; that is to say, if you use 100 g masses then the toy car will accelerate very quickly and manual timing will prove to be impossible. In practice, we found that adding small metal washers to an improvised hook made from a paper clip worked well. We found the average mass of the washers by placing ten of them on a scale.

Then input the data into this spreadsheet (click the link to download from Google Drive) and the software should do the rest (including plotting the graph!).

The Eleventh Commandment: Thou Shalt Not Confound Thy Variables!

To confirm the straight line and directly proportional relationship between accelerating force and acceleration, bear in mind that the total mass of the accelerating system must remain constant in order for it to be a “fair test”.

The parts of our system that are accelerating are the toy car, the string and the weight stack. The total mass of the accelerating system shown below is 461 g (assuming the mass of the hook and the string are negligible).

The accelerating (or resultant) force is the weight of 0.2 g mass on the hook, which0 can be calculated using W = mg and will be equal to 0.00196 N or 1.96 mN.

In the second diagram, we have increased the mass on the weight stack to 0.4 g (and the accelerating force to 0.00392 N or 3.92 mN) but note that the total mass of the accelerating system is still the same at 461 g.

In practice, we found that using blu-tac to stick a matchbox tray to the roof of the car made managing and transferring the weight stack easier.

Personal note: as a beginning teacher, I demonstrated the linear air track version of this experiment to an A-level Physics class and ended up disconfirming Newton’s Second Law instead of confirming it; I was both embarrassed and immensely puzzled until an older, wiser colleague pointed out that the variables had been well and truly confounded by not keeping the total mass of the accelerating system constant.

It was embarrassing and that’s why I always harp on about this aspect of the experiment.

What lies beneath: the Physics underlying this method

This can be considered as “deep background” rather than necessary information, but I, for one, consider it really interesting.

Acceleration is the rate of change of a rate of change. Velocity is the rate of change of displacement with time and acceleration is the rate of change of velocity.

Interested individuals may care to delve into higher derivatives like jerk, snap, crackle and pop (I kid you not — these are the technical terms). Jerk is the rate of change of acceleration and hence can be defined as (takes a deep breath) the rate of change of a rate of change of rate of change. More can be found in the fascinating article by Eager, Pendrill and Reistad (2016) linked to above.

But on a much more prosaic level, acceleration can be defined as a = (v – u) / t where v is the final instantaneous velocity, u is the inital instantaneous velocity and t is the time taken for the change.

The instantaneous velocity is the velocity at a momentary instant of time. It is, if you like, the velocity indicated by the needle on a speedometer at a single instant of time and is different from the average velocity which is calculated from the total distance travelled divided the time taken.

This can be shown in diagram form like this:

However, our experiment is simplified because we made sure that the toy car was stationary when the timer was zero; in other words, we ensured u = 0 m/s.

This simplifies a = (v – u) / t to a = v / t.

But how can we find v, the instantaneous velocity at the end of the journey when we have no direct means of measuring it, such as a speedometer or a light gate?

No more jerks left to give

Let’s assume that, for the toy car, the jerk is zero (again, let me emphasize that jerk is a technical term defined as the rate of change of acceleration).

This means that the acceleration is constant.

This fact allows us to calculate the average velocity using a very simple formula: average velocity = (u + v) / t .

But remember that u = 0 so average velocity = v / 2 .

More pertinently for us, provided that u = 0 and jerk = 0, it allows us to calculate a value for v using v = 2 x (average velocity) .

The spreadsheet linked to above uses this formula to calculate v and then uses a = v / t.

Using this in the school laboratory

This could be done as a demonstration or, since only basic equipment is needed, a class experiment. Students may need access to computers running the spreadsheet during the experiment or soon afterwards. We found that one laptop shared between two groups was sufficient.

First experiment (relationship between force and acceleration): set up as shown in the diagram. Place washers totalling a mass of 0.8 g (or similar) and washers totalling a mass of 0.2 g on the hook or weight stack. Hold the toy car stationary at the start line. Release and start the timer. Stop the timer. Input data into the spreadsheet and repeat with different mass on the hook.

It can be useful to get students to manually “check” the value of a calculated by the spreadsheet to provide low stakes practice of using the acceleration formula.

Second experiment (relationship between mass and acceleration). Keep the accelerating force constant with (say) 0.6 g on the hook or weight stack. Hold the toy car stationary at the start line. Release and start the timer. Stop the timer. Input data into the second tab on the spreadsheet and repeat with 100 g added to the toy car (possibly blu-tac’ed into place).

Conclusion

This blog post grew in the telling. Please let me know if you try the methods outlined here and how successful you found them

References

Eager, D., Pendrill, A. M., & Reistad, N. (2016). Beyond velocity and acceleration: jerk, snap and higher derivatives. European Journal of Physics, 37(6), 065008.

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