The ‘all-in-a-row’ circuit diagram convention for series and parallel circuits

Circuit diagrams can be seen either as pictures or abstractions but it is clear that pupils often find it hard to recognise the circuits in the practical situation of real equipment. Moreover, Caillot found that students retain from their work with diagrams strong images rather than the principles they are intended to establish. The topological arrangement of a diagram or a drawing presents problems for pupils which are easily overlooked. It seems that pupils’ spatial abilities affect their use of circuit diagrams: they sometimes do not regard as identical several circuits, which, though identical, have been rotated so as to have a different spatial arrangement. […] Niedderer found that pupils, when asked whether a circuit diagram would ‘work’ in practice, more often judged symmetrical diagrams to be functioning than non-symmetric ones.

Driver et al. (1994): 124 [Emphases added]

For the reasons outlined by Driver and others above, I think it’s a good idea to vary the way that we present circuit diagrams to students when teaching electric circuits. If students always see circuit diagrams presented so that (say) the cell is at the ‘top’ and ‘facing’ a certain way; or that they are drawn so that they are symmetrical (which is an aesthetic rather that a scientific choice), then they may well incorrectly infer that these and other ‘accidental’ features of our circuit diagrams are the essential aspects that they should pay the most attention to.

One ‘shake it up’ strategy is to redraw a circuit diagram using the ‘all-in-a-row’ convention.

If you arrange the real components in the ‘all-in-a-row’ arrangement, then a standard digital voltmeter has, what is in my opinion a regrettably underused functionality, that will show:

  • ‘positive’ potential differences: that is to say, the energy added to the coulombs as they pass through a cell or the electromotive force; and
  • ‘negative’ potential differences: that is to say, the energy removed from each coulomb as they pass through a resistor; these can be usefully referred to as ‘potential drops’

This can be shown on circuit diagrams as shown below/

In other words, the difference between the potential difference across the cell (energy being transferred into the circuit from the chemical energy store of the cell) is explicitly distinguished from the potential difference across the resistor (energy being transferred from the resistor into the thermal energy store of the surroundings). The all-in-a-row convention neatly sidesteps a common misconception that the potential difference across a cell is equal to the potential difference across a resistor: they are not. While they may be numerically equal, they are different in sign, as a consequence of Kirchoff’s Second Law. As I have suggested before, I think that this misconception is due to the ‘hidden rotation‘ built into standard circuit diagrams.

Potential divider circuits and the all-in-a-row convention

Although I am normally a strong proponent of the ‘parallel first heresy‘, I’ll go with the flow of ‘series circuit first’ in this post.

Diagrams 2 and 3 in the sequence show that the energy supplied to the coulombs (+1.5 V or 1.5 joules per coulomb) by the cell is transferred from the coulombs as they pass through the double resistor combination. Assuming that R1 = R2 then, as diagram 4 shows, 0.75 joules will be transferred out of each coulomb as they pass through R1; as diagram 5 shows, 0.75 joules will be transferred out of each coulomb as they pass through R2.

Parallel circuits and the all-in-a-row convention

I’ve written about using the all-in-a-row convention to help explain current flow in parallel circuits here, so I will focus on understanding potential difference in parallel circuit in this post.

Again, diagrams 2 and 3 in the sequence show that the positive 3.0 V potential difference supplied by the cell is numerically equal (but opposite in sign) to the negative 3.0 V potential drop across the double resistor combination. It is worth bearing in mind that each coulomb passing through the cell gains 3.0 joules of energy from the chemical energy store of the cell. Diagrams 4 and 5 show that each coulomb passing through either R1 or R1 loses its entire 3.0 joules of energy as it passes through that resistor. The all-in-a-row convention is useful, I think, for showing that each coulomb passes through just one resistor as it makes a single journey around the circuit.

Acknowledgements

Circuit simulations from the excellent https://phet.colorado.edu/sims/html/circuit-construction-kit-dc/latest/circuit-construction-kit-dc_en.html

Circuit diagrams drawn using https://www.circuit-diagram.org/editor/

Reference

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

Electromagnetic induction — using the LEFT hand rule…?

They do observe I grow to infinite purchase,
The left hand way;

John Webster, The Duchess of Malfi

Electromagnetic induction — the fact that moving a conductor inside a magnetic field in a certain direction will generate (or induce) a potential difference across its ends — is one of those rare-in-everyday-life phenomena that students very likely will never have come across before. In their experience, potential differences have heretofore been produced by chemical cells or by power supply units that have to be plugged into the mains supply. Because of this, many of them struggle to integrate electromagnetic induction (EMI) into their physical schema. It just seems such a random, free floating and unconnected fact.

What follows is a suggested teaching sequence that can help GCSE-level students accept the physical reality of EMI without outraging their physical intuition or appealing to a sketchily-explained idea of ‘cutting the field lines’.

‘Look, Ma! No electrical cell!’

I think it is immensely helpful for students to see a real example of EMI in the school laboratory, using something like the arrangement shown below.

A length of copper wire used to cut the magnetic field between two Magnadur magnets on a yoke will induce (generate) a small potential difference of about 5 millivolts. What is particularly noteworthy about doing this as a class experiment is how many students ask ‘How can there be a potential difference without a cell or a power supply?’

The point of this experiment is that in this instance the student is the power supply: the faster they plunge the wire between the magnets then the larger the potential difference that will be induced. Their kinetic energy store is being used to generate electrical power instead of the more usual chemical energy store of a cell.

But how to explain this to students?

A common option at this point is to start talking about the conductor cutting magnetic field lines: this is hugely valuable, but I recommend holding fire on this picture for now — at least for novice learners.

What I suggest is that we explain EMI in terms of a topic that students will have recently covered: the motor effect.

This has two big ‘wins’:

  • It gives a further opportunity for students to practice and apply their knowledge of the motor effect.
  • Students get the chance to explain an initially unknown phenomenon (EMI) in terms of better understood phenomenon (motor effect). The motor effect will hopefully act as the footing (to use a term from the construction industry) for their future understanding of EMI.

Explaining EMI using the motor effect

The copper conductor contains many free conduction electrons. When the conductor is moved sharply downwards, the electrons are carried downwards as well. In effect, the downward moving conductor can be thought of as a flow of charge; or, more to the point, as an electrical current. However, since electrons are negatively charged, this downward flow of negative charge is equivalent to an upward flow of positive charge. That is to say, the conventional current direction on this diagram is upwards.

Applying Fleming Left Hand Rule (FLHR) to this instance, we find that each electron experiences a small force tugging it to the left — but only while the conductor is being moved downwards.

This results in the left hand side of the conductor becoming negatively charged and the right hand side becoming positively charged: in short, a potential difference builds up across the conductor. This potential difference only happens when the conductor is moving through the magnetic field in such a way that the electrons are tugged towards one end of the conductor. (There is, of course, the Hall Effect in some other instances, but we won’t go into that here.)

As soon as the conductor stops moving, the potential difference is no longer induced as there is no ‘charge flow’ through the magnetic field and, hence, no current and no FLHR motor effect force acting on the electrons.

Faraday’s model of electromagnetic induction

Michael Faraday (1791-1867) discovered the phenomenon of electromagnetic induction in 1831 and explained it using the idea of a conductor cutting magnetic field lines. This is an immensely valuable model which not only explains EMI but can also generate quantitative predictions and, yes, it should definitely be taught to students — but perhaps the approach outlined above is better to introduce EMI to students.

The left hand rule not knowing what the right hand rule is doing . . .

We usually apply Fleming’s Right Hand Rule (FRHW) to cases of EMI, Can we replace its use with FLHR? Perhaps, if you wanted to. However, FRHR is a more direct and straightforward shortcut to predicting the direction of conventional current in this type of situation.

Split ring commutator? More like split ring commuHATER!

Students find learning about electric motors difficult because:

  1. They find it hard to predict the direction of the force produced on a conductor in a magnetic field, either with or without Fleming’s Left Hand Rule.
  2. They find it hard to understand how a split ring commutator works.

In this post, I want to focus on a suggested teaching sequence for the action of a split ring commutator, since I’ve covered the first point in previous posts.

Who needs a ‘split ring commutator’ anyway?

We all do, if we are going to build electric motors that produce a continuous turning motion.

If we naively connected the ends of a coil to power supply, then the coil would make a partial turn and then lock in place, as shown below. When the coil is in the vertical position, then neither of the Fleming’s Left Hand Rule (FLHR) forces will produce a turning moment around the axis of rotation.

When the coil moves into this vertical position, two things would need to happen in order to keep the coil rotating continuously in the same direction.

  • The current to the coil needs to be stopped at this point, because the FLHR forces acting at this moment would tend to hold the coil stationary in a vertical position. If the current was cut at this time, then the momentum of the moving coil would tend to keep it moving past this ‘sticking point’.
  • The direction of the current needs to be reversed at this point so that we get a downward FLHR force acting on side X and an upward FLHR force acting on side Y. This combination of forces would keep the coil rotating clockwise.

This sounds like a tall order, but a little device known as a split ring commutator can help here.

One (split) ring to rotate them all

The word commutator shares the same root as commute and comes from the Latin commutare (‘com-‘ = all and ‘-mutare‘ = change) and essentially means ‘everything changes’. In the 1840s it was adopted as the name for an apparatus that ‘reverses the direction of electrical current from a battery without changing the arrangement of the conductors’.

In the context of this post, commutator refers to a rotary switch that periodically reverses the current between the coil and the external circuit. This rotary switch takes the form of a conductive ring with two gaps: hence split ring.

Tracking the rotation of a coil through a whole rotation

In this picture below, we show the coil connected to a dc power supply via two ‘brushes’ which rest against the split ring commutator (SRC). Current is flowing towards us through side X of the coil and away from us through side Y of the coil (as shown by the dot and cross 2D version of the diagram. This produces an upward FLHR force on side X and a downward FLHR force on side Y which makes the coil rotate clockwise.

Now let’s look at the coil when it has turned 45 degrees. We note that the SRC has also turned by 45 degrees. However, it is still in contact with the brushes that supply the current. The forces on side X and side Y are as noted before so the coil continues to turn clockwise.

Next, we look at the situation when the coil has turned by another 45 degrees. The coil is now in a vertical position. However, we see that the gaps in the SRC are now opposite the brushes. This means that no current is being supplied to the coil at this point, so there are no FLHR forces acting on sides X and side Y. The coil is free to continue rotating clockwise because of momentum.

Let’s now look at the situation when the coil has rotated a further 45 degrees to the orientation shown below. Note that the side of the SRC connected to X is now touching the brush connected to the positive side of the power supply. This means that current is now flowing away from us through side X (whereas previously it was flowing towards us). The current has reversed direction. This creates a downward FLHR force on side X and an upward FLHR force on side Y (since the current in Y has also reversed direction).

And a short time later when the coil has moved a total of180 degrees from its starting point, we can observe:

And later:

And later still:

And then:

And then eventually we get back to:

Summary

In short, a split ring commutator is a rotary switch in a dc electric motor that reverses the current direction through the coil each half turn to keep it rotating continuously.

A powerpoint of the images used is here:

And a worksheet that students can annotate (and draw the 2D versions of the diagrams!) is here:

I hope that this teaching sequence will allow more students to be comfortable with the concept of a split ring commutator — anything that results in a fewer split ring commuHATERS would be a win for me 😉

Explaining current flow in conductors (part three)

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.

In part two we looked at how the distribution of surface charges on electrical conductors produces the internal electric fields that guide and push charges around electric circuits and highlighted the published evidence that supports this model.

A representation of a surface charge distribution giving rise to the internal electric field (purple arrows) of a current-carrying conductor

In part three, we are going to look at the transient processes that produce the required distribution of surface charges. In this treatment, I am going to lean very heavily on the analysis presented in Duffin (1980: 167-8).

Connecting wires to a chemical cell

Let’s connect up a simple circuit using a chemical cell as our source of EMF ℰ.

The first diagram shows the cell and wires before they are connected.

When the wires are connected there is a momentary current flow from the cell that creates the surface charge distribution shown below.

The current will stop when the ends of the wire at a potential difference V which is equal to the EMF ℰ of the cell. The ends of the wire act as a small capacitor (∼10-15 F or less). The wires act as equipotential volumes so the very small charge must be distributed over the surface of the wires with a slight concentration of charge at the ends.

Making the circuit

If the ends of the wire are now connected, then the capacitance drops to zero and the ends of the wires become discharged. This leads to very low concentration of surface charge in this region.

However, just enough surface charge remains to produce the internal electric field as shown below. The field lines of the internal electric field are parallel to the wire.

The potential diagram is after Figure 6.17 (Duffin 1980: 160). The ‘dip’ between C and A is due to the effect of the internal resistance of the cell. As we can see in this instance, when there is a steady flow of current then V is slightly smaller than ℰ.

Reference

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

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

Next post

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

References

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

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

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

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

Explaining current flow in conductors (part one)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Understanding the cause of current flow

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

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

Härtel (2021): 21

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

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

References

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

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

Acknowledgements

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

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

Modelling Electrical Power Equations using the CTM

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

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

E=QV and the CTM

An animated version of the Coulomb Train Model

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

Using the Coulomb Train Model:

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

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

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

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

Hewitt representation of increasing Q and V simultaneously

All you need is E = Q V

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

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

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

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

Deriving P=I2R from P=IV

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

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

We can increase the power transferred to the resistor by:

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

Linking the electrical power formulas

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

Teaching Circuits using Kirchoff’s Laws

Gustav Robert Kirchhoff (1824 – 1887) was a pioneer in the study of the radiation given off by hot objects and was the first person to use the term ‘black body radiation’. He also made groundbreaking contributions to what was then the ‘new’ science of spectroscopy.

High school students first encounter his name when studying electric circuits. Kirchhoff developed laws which describe the behaviour of electric circuits and, rightly, these laws still bear his name.

Newton needed three laws to explain the whole of motion; Kirchhoff needed only two to explain the behaviour of all circuits.

Kirchhoff’s First Law (aka KCL or Kirchhoff’s Current Law)

This law is a consequence of the Principle of Conservation of Electric Charge.

The algebraic sum of all the currents flowing through all the wires in a network that meet at a point is zero

Oxford Dictionary of Physics (2015)

‘Algebraic sum’ means that we must take account of whether the electric currents are positive or negative; or, in other words, their direction.

This can be stated more simply as: the sum of electric currents flowing into a junction is equal to the sum of the electric currents flowing out of the junction.

Kirchhoff’s Second Law (aka KVL or Kirchhoff’s Voltage Law)

This law is a consequence of the Principle of Conservation of Energy.

The algebraic sum of the e.m.f.s within any closed circuit is equal to the sum of the products of the currents and the resistances in the various portions of the circuit.

Oxford Dictionary of Physics (2015)

This can be more directly understood as saying that in any closed loop of the circuit, the sum of the energies gained by the charge carriers as they pass through parts of the loop with a positive potential difference is equal to the sum of the energies lost by the charge carriers as they pass through parts of the circuit with electrical resistance.

In other words, the sum of the positive potential differences (e.m.f.s) is equal to the sum of the negative potential differences around any closed loop of a circuit.

Applying Kirchhoff’s Laws to a circuit problem

Find the current flowing through each resistor in this circuit

Applying Kirchhoff’s First Law

But wait — is the current I2 flowing in the right way? Surely it should be coming out of the positive terminal of the 7 V battery, no?

Perhaps. The direction of I2 is simply my semi-educated guess at its direction given the relative strength of the 27 V cell and the 7 V cell.

But the happy truth of using Kirchhoff’s First Law is that . . . it doesn’t matter. Even if we have guessed the direction wrong, after we have gone through the process all that will happen is that we will get the correct numerical value for I2 but our mistake will be revealed by the fact that it will have a negative value.

If we look at the junction highlighted in yellow, we can see that the algebriac sum of currents is: I1 – I2 – I3 = 0.

We can rewrite this as: I1 = I2 + I3.

And that’s about as far as we can get using just Kirchhoff’s First Law. We have three unknowns so somehow we need to obtain two more independent expressions of their relationships to solve this circuitous conundrum.

Luckily, we still have Kirchhoff Second Law to bring into play…

Applying Kirchhoff’s Second Law (Part 1 of 2)

Before starting, I find it immensely helpful to indicate which ends of the components have a positive potential and which have a negative potential. This process is part of a general maxim that I try to apply to all areas of physics problem solving: why think hard when your diagram can do the thinking for you? (See here for a similar process applied to dynamics problems.)

An example of ‘Why think hard when the diagram will do it for you?’

Going around loop L1 in the direction shown by the arrow:

  • The emf 27 V will be positive as the potential increases (as we are moving from – to +).
  • I3R3 will be negative as the potential decreases (as we are moving from + to -).
  • I1R1 will be negative as the potential decreases (as we are moving from + to -).

This gives us:

Applying Kirchhoff’s Second Law (Part 2 of 2)

Yet another example of ‘Let the diagram do the hard thinking for you!’

Going around Loop L2 we find that:

  • The emf 7 V will be positive as the potential increases (as we are moving from – to +).
  • I2R2 will be positive as the potential increases (as we are moving from – to +).
  • I3R3 will be negative as the potential decreases (as we are moving from + to -).

This gives us:

The rest is history algebra

Going through a rather involved process of using simultaneous equations for solving for I1, I2 and I3 . . .

(NB There’s probably a quicker way than the way I chose, but I got there in the end and that’s the important thing. AND I guessed the direction of I2 correctly. *Pats himself on the back*).

Conclusion

Kirchhoff’s Laws: I hope you give them a spin, whether or not you decide to use the procedure outlined above 🙂

Series and Parallel Circuits — an unhelpful dichotomy?

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

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

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

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

The actual situation is more like this:

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

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

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

How to avoid the false dichotomy

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Conclusion

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

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

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

Circuit Diagrams: Lost in Rotation…?

Is there a better way of presenting circuit diagrams to our students that will aid their understanding of potential difference?

I think that, possibly, there may be.

(Note: circuit diagrams produced using the excellent — and free! — web editor at https://www.circuit-diagram.org/.)

Old ways are the best ways…? (Spoiler: not always)

This is a very typical, conventional way of showing a simple circuit.

A simple circuit as usually presented

Now let’s measure the potential difference across the cell…

Measuring the potential difference across the cell

…and across the resistor.

Measuring the potential difference across the resistor

Using a standard school laboratory digital voltmeter and assuming a cell of emf 1.5 V and negligible internal resistance we would get a value of +1.5 volts for both positions.

Let me demonstrate this using the excellent — and free! — pHET circuit simulation website.

Indeed, one might argue with some very sound justification that both measurements are actually of the same potential difference and that there is no real difference between what we chose to call ‘the potential difference across the cell’ and ‘the potential difference across the resistor’.

Try another way…

But let’s consider drawing the circuit a different (but operationally identical) way:

The same circuit drawn ‘all-in-a-row’

What would happen if we measured the potential difference across the cell and the resistor as before…

This time, we get a reading (same assumptions as before) of [positive] +1.5 volts of potential difference for the potential difference across the cell and [negative] -1.5 volts for the potential difference across the resistor.

This, at least to me, is a far more conceptually helpful result for student understanding. It implies that the charge carriers are gaining energy as they pass through the cell, but losing energy as they pass through the resistor.

Using the Coulomb Train Model of circuit behaviour, this could be shown like this:

+1.5 V of potential difference represented using the Coulomb Train Model
-1.5 V of potential difference represented using the Coulomb Train Model. (Note: for a single resistor circuit, the emerging coulomb would have zero energy.)

We can, of course, obtain a similar result for the conventional layout, but only at the cost of ‘crossing the leads’ — a sin as heinous as ‘crossing the beams’ for some students (assuming they have seen the original Ghostbusters movie).

Crossing the leads on a voltmeter

A Hidden Rotation?

The argument I am making is that the conventional way of drawing simple circuits involves an implicit and hidden rotation of 180 degrees in terms of which end of the resistor is at a more positive potential.

A hidden rotation…?

Of course, experienced physics learners and instructors take this ‘hidden rotation’ in their stride. It is an example of the ‘curse of knowledge’: because we feel that it is not confusing we fail to anticipate that novice learners could find it confusing. Wherever possible, we should seek to make whatever is implicit as explicit as we can.

Conclusion

A translation is, of course, a sliding transformation, rather than a circumrotation. Hence, I had to dispense with this post’s original title of ‘Circuit Diagrams: Lost in Translation’.

However, I do genuinely feel that some students understanding of circuits could be inadvertently ‘lost in rotation’ as argued above.

I hope my fellow physics teachers try introducing potential difference using the ‘all-in-row’ orientation shown.

The all-in=a-row orientation for circuit diagrams to help student understanding of potential difference

I would be fascinated to know if they feel its a helpful contribition to their teaching repetoire!