Physics › Electricity › Current, charge and the direction problem

Current, charge and the direction problem

Three definitions carry the whole of electricity. Current is a rate of flow of charge. Potential difference is the energy each coulomb picks up or hands over. Divide one by the other and you have resistance. Alongside them runs a historical accident, because the arrows on every circuit diagram mark the defined direction of positive-charge flow, and in a metal the electrons drift the other way.

Year 12AQA 3.5.1.1 CIE 9.1.1, 9.1.2, 9.1.3, 9.1.4, 9.2.1, 9.2.2, 9.3.1, 9.3.2 OCR A 4.1.1(a), 4.1.1(b), 4.1.1(c), 4.1.1(d), 4.1.1(e), 4.1.1(f), 4.1.2(a), 4.1.2(b), 4.1.2(c), 4.2.2(a), 4.2.2(d), 4.2.2(e), 4.2.3(a)EDEXCEL 31, 32, 41

IN THIS TOPIC

Use I = ΔQ/Δt, treating the coulomb as an amp second.
Name the carriers in each case: delocalised electrons in a metal, ions of both signs in an electrolyte.
Get energy out of V = W/Q, where one volt means one joule per coulomb.
Send a charge across a pd through empty space and calculate the result with eV = ½mv².
R = V/I is a definition. State it as one, and keep conventional current apart from electron drift.
Connect a current to the carrier density and drift speed behind it with I = nAvq.

COMMON MISCONCEPTION

The current arrows show which way the electrons go.

Current is a rate

Electric current is the rate of flow of charge.

I = \frac{\Delta Q}{\Delta t}

ON THE AQA DATA SHEET

Stand at one cross-section of a wire and count the charge passing per second. That count is the current. One amp is one coulomb per second, so a coulomb is an amp second. Rearranged, ΔQ = IΔt, and that is the form nearly every calculation actually uses.

FIG. 1Electrons drift left along the wire, and at this pace one of the drawn cohort crosses the counting line every second and a half. Real current is the total charge crossing per second, unimaginably many electrons at once rather than the handful a drawing can show. Conventional current points the other way, a bookkeeping choice made before anyone knew what moved. The real drift in copper is a fraction of a millimetre per second, slowed further here so you can see it.

Charge itself comes in fixed lumps. Every electron carries a charge of magnitude 1.60 × 10⁻¹⁹ C, negative in sign, a constant printed in the data booklet, and any measured charge is a whole number of those lumps. That is how a question turns coulombs into a count of electrons.

A current needs carriers, and which particles do the carrying depends on the material. A metal is a lattice of fixed positive ions sitting in a sea of delocalised electrons, loose from any one atom and free to drift through the whole structure. A current in a wire is that electron population on the move, and nothing else in the metal travels at all.

An electrolyte conducts by moving matter instead. Melt an ionic solid or dissolve one in water, copper sulfate solution being the standard case, and the ions themselves come free of each other. Both signs then drift, in opposite directions: positive ions towards the negative electrode, negative ions towards the positive one. The two flows add rather than cancel, because negative charge going one way counts the same as positive charge going the other, and the total is the current the ammeter in the leads reads. Electrons carry it through the wires, ions carry it through the liquid, and the same coulombs per second pass through each.

That difference in carrier shows up as a difference you can weigh. Ions are matter, so a current through an electrolyte transports substance as well as charge, plating copper onto one electrode and stripping it off the other. A current through a copper wire moves no copper anywhere.

The direction problem

In the 1750s Benjamin Franklin named the two kinds of charge long before anyone knew what moved in a wire. The electron was not discovered for another 140 years, and it turned out to carry the charge his scheme calls negative, so in a metal the moving particles drift against the marked direction. By then every rule and diagram had been built on the convention, and physics kept it. Nothing about that is wrong: conventional current is the defined direction of positive-charge flow, and it works for every circuit law, whatever the carriers happen to be.

FIG. 2Conventional current runs from the positive terminal to the negative; the electrons drift the opposite way along the same wire.

So conventional current runs from + to − around the outside of a circuit, and that is the direction every arrow and every rule in the subject uses. The electrons, the carriers actually moving in a metal, drift from − to +, backwards. Both statements are true at once; keep the labels attached and nothing goes wrong.

The electrolyte is the place where the marked direction and the moving particles agree. Inside the liquid the positive ions drift towards the negative electrode, which is the direction the current arrow claims, and the negative ions drifting the opposite way count the same way round. It is the metal, with nothing free to move in it but electrons, where the carriers travel against the marked direction.

One more indignity for the electrons. They drift at a fraction of a millimetre per second, slower than a snail, and yet the lights come on the moment you flick the switch. The push travels round the circuit at close to the speed of light. The signal is fast; the particles are not.

Potential difference

Potential difference tracks the energy. It is the work done per unit charge.

V = \frac{W}{Q}

ON THE AQA DATA SHEET

The volt is therefore a joule per coulomb. A 1.5 V cell gives each coulomb 1.5 J on the way through; a component with 1.5 V across it takes 1.5 J back off each coulomb that passes.

FIG. 3Around the loop, each coulomb collects its energy at the cell and hands it over at the component. Potential difference measures the joules per coulomb at each stage.

Rearranged, W = VQ, and with Q = It that becomes W = VIt. Those two lines are the road from electrical quantities to joules, and every energy calculation in the unit travels it.

None of that depends on whether the charge is threading a wire. Send one charged particle across the same pd through empty space and the work done on it is still QV, except that now there is no lattice to hand any of it to. All of it becomes kinetic energy. For an electron of charge e released from rest and accelerated through a pd V, that reads $eV = \frac{1}{2}mv^{2}$ , and for any other charged particle q takes the place of e. Three conditions come with it: a vacuum, a start from rest, and speeds low enough for Newtonian mechanics to hold. Inside those, it is the working equation of every electron gun and the first stage of every accelerator.

WORKED EXAMPLE

The speed out of an electron gun

An electron starts from rest and crosses a pd of 2.0 kV in a vacuum. Find its speed. (e = 1.60 × 10⁻¹⁹ C, m_e = 9.11 × 10⁻³¹ kg.)

Every joule of work becomes kinetic energy, so eV = ½mv² and v = $\sqrt{2eV/m}$ .

v = √(2 × 1.60 × 10⁻¹⁹ × 2000 / (9.11 × 10⁻³¹)) = 2.7 × 10⁷ m s⁻¹.

Nine per cent of the speed of light out of a bench supply, and something like 10¹¹ times the drift speed of the electrons in the wire that feeds the gun. Same pd, same energy per coulomb; the difference is that in the wire the charge keeps colliding with the lattice and giving the energy straight back.

WORKED EXAMPLE

From current to charge to energy

A 6.0 V battery drives 0.50 A through a lamp for one minute. Find the charge that passes and the energy transferred.

Charge first. Q = It = 0.50 × 60 = 30 C, with the minute written as 60 seconds. That conversion is where this question sheds most of its marks.

Now the energy. W = VQ = 6.0 × 30 = 180 J.

Read the numbers back as a sentence. Thirty coulombs went round the loop, the battery handed each of them six joules, and each spent them in the lamp.

The drift equation

How does a current of whole amps flow when each electron crawls along at well under a millimetre per second? Sheer numbers. A copper wire holds around 10²⁹ free electrons in every cubic metre, and a current is that entire population creeping together. The link between the crawl and the amps is

I = nAvq

where n is the number density of charge carriers, the count per cubic metre, A is the cross-sectional area of the wire, v is the drift speed and q is the charge on each carrier. The logic takes one sentence. In one second every carrier within a distance v of a chosen cross-section reaches it, that slab of wire has volume Av and holds nAv carriers, so the charge passing per second is nAvq.

WORKED EXAMPLE

The famous crawl, derived

A copper wire of cross-sectional area 1.0 mm² carries 5.0 A. Copper has n = 8.5 × 10²⁸ m⁻³. Find the drift speed.

Rearrange to v = I/nAq, and put the area into square metres, A = 1.0 × 10⁻⁶ m².

v = 5.0 / (8.5 × 10²⁸ × 1.0 × 10⁻⁶ × 1.60 × 10⁻¹⁹) = 3.7 × 10⁻⁴ m s⁻¹.

A third of a millimetre per second, the slow drift described earlier. The current is respectable because n is astronomical, not because anything moves quickly.

Three everyday facts follow from the same equation. Halve the area and the drift speed has to double to carry the current. A bulb's filament runs white-hot while its leads stay cool because one current threads every part of a series path, so the power I²R piles up wherever R is largest, and a long thin strand of high-resistivity alloy is exactly that place. In a semiconductor n is smaller by something like a factor of a billion, so the same current demands drift speeds a billion times greater.

Resistance, defined

Resistance is defined as the pd across a component divided by the current through it.

R = \frac{V}{I}

ON THE AQA DATA SHEET

Its unit is the ohm, one volt per amp. That is a definition. No law is being claimed, and nothing in it requires the ratio to stay constant when conditions change. For most components it does not. Sorting out which components hold the ratio fixed occupies the whole of the next lesson.

GUIDED PRACTICE

A heater's resistance

A mains heater draws 8.5 A from a 230 V supply. Find its resistance from the definition.

Show the working

R = V/I = 230/8.5 = 27 Ω.

A ratio taken at one operating point is all you have, and this one is the hot resistance. Measure the same element cold and it reads far lower.

INDEPENDENT PRACTICE

A lightning bolt's current

A lightning stroke transfers about 5 C of charge in roughly 100 μs. Estimate the current.

Show the working

I = Q/t = 5/(1.0 × 10⁻⁴) = 5 × 10⁴ A.

Fifty thousand amps, out of the same I = Q/t that handles a torch bulb. The definition scales without complaint. A current that size heats the air channel until it blows outwards, and that is thunder.

ASSESSMENT FOCUS

Time goes into ΔQ = IΔt in seconds. An unconverted minute is the classic first-line error, and it takes the rest of the answer down with it.
“How many electrons?” means divide the charge by 1.60 × 10⁻¹⁹ C. Expect something between 10¹⁹ and 10²¹. A few dozen means the powers of ten went astray.
Label every direction you state. Conventional current runs + to − and electron flow runs − to +, and a bare arrow lets the examiner read it whichever way suits.
Write the definitions as sentences. Current is the rate of flow of charge, pd is the work done per unit charge, and both come up in words as often as in symbols.
W = VQ, or W = VIt when a time is given. Pick whichever matches the quantities already on the page.
Asked what is actually moving, name the carrier for the material in front of you. Delocalised electrons in a metal, and ions of both signs drifting in opposite directions in an electrolyte, their two contributions adding. Answering “electrons” for the electrolyte gives away a mark that was there for the taking.
eV = ½mv² comes with conditions, and stating them is part of the answer. From rest, in a vacuum, with all the work becoming kinetic energy. Swap in qV for a particle that is not an electron, and keep the accelerating pd out of any resistance calculation, since nothing along the way is dissipated.
AQA never prints I = nAvq, though OCR A and CIE both use it. Say what n means physically as well as substituting into it, since the sentence carries a mark of its own.

CHECK YOURSELF

A lamp carries a current of 0.25 A for 2.0 minutes. (a) How much charge passes through it? (b) How many electrons is that?

Show a hint

Seconds first. Then remember that charge comes in fixed lumps.

Show the answer

(a) $\Delta Q = I\Delta t$ = 0.25 × 120 = 30 C.

(b) Each electron carries 1.60 × 10⁻¹⁹ C, so N = 30 / (1.60 × 10⁻¹⁹) = 1.9 × 10²⁰ electrons. A colossal number for a modest lamp. At counts like that, grainy charge behaves as a smooth fluid.

Conventional current runs plus to minus.

The electrons go the other way, slowly.

Both are true at once. Keep the labels on.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

20 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the current, charge and the direction problem questions page.

8 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

CHECK YOUR PROGRESS

Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.

Use I = ΔQ/Δt, treating the coulomb as an amp second.
Name the carriers in each case: delocalised electrons in a metal, ions of both signs in an electrolyte.
Get energy out of V = W/Q, where one volt means one joule per coulomb.
Send a charge across a pd through empty space and calculate the result with eV = ½mv².
R = V/I is a definition. State it as one, and keep conventional current apart from electron drift.
Connect a current to the carrier density and drift speed behind it with I = nAvq.

Open the full revision checklist to track your progress across the whole unit.

Current, charge and the direction problem

Current is a rate

The direction problem

Potential difference

The drift equation

Resistance, defined

Question workbook

Question workbook

Answer book