PhysicsCore22 min read

Electricity & Magnetism

Charge, current, circuits and fields

This topic appears in:

01

Charge, and why only electrons move

Definition

Electric charge — A property of matter that causes it to experience a force in an electric field. It comes in two kinds, positive and negative, and is measured in coulombs.

Like charges repel and unlike charges attract. That single rule explains everything from a balloon sticking to a wall to the structure of the atom itself.

In a solid, the positive charges are the protons, locked inside nuclei that are fixed in the structure of the material. They cannot go anywhere. Only the electrons are free to move, and in a metal a few of them are very free indeed — detached from their parent atoms and able to drift through the whole lattice.

This is why an object becomes positively charged by losing electrons, never by gaining protons. Rub a polythene rod with a cloth and electrons transfer from cloth to rod: the rod becomes negative, the cloth positive. Saying that "positive charge moved to the cloth" describes the same outcome but scores nothing, because it is not what happened.

Charge is also quantised. Every charge you can measure is a whole-number multiple of the elementary charge e = 1.6 × 10⁻¹⁹ C. There is no such thing as half an electron of charge.

Conductors and insulators

A conductor has free electrons that can drift through it — metals, and graphite. An insulator has none: every electron is bound to its atom, so charge placed on it stays exactly where you put it. That is why a charged plastic rod holds its charge and a charged metal rod loses it to your hand instantly.

02

Current is charge on the move

Current is the rate at which charge flows past a point. One ampere means one coulomb passing every second — roughly six million million million electrons.

Two things about current cause endless confusion, and both are worth settling now. First, current is not used up going round a circuit. The ammeter reading before a lamp and after it are identical. What the lamp takes is energy, not charge.

Second, conventional current is defined as flowing from the positive terminal of a supply to the negative. Electrons actually drift the opposite way, because they are negative. Every rule you will use — the motor rule, the right-hand grip rule, the direction arrows on a circuit diagram — is stated for conventional current. Use it consistently and the electrons' disagreement never matters.

I = Q / tQ = I tone ampere is one coulomb per second
I
currentA
Q
chargeC
t
times

Raise the resistance and watch the current fall for the same supply voltage. The relationship is I = V/R — halving the resistance doubles the current.

03

Potential difference and e.m.f.

Definition

Potential difference — The energy transferred from each coulomb of charge as it passes through a component, V = E/Q. One volt is one joule per coulomb.

A battery does not supply charge — the charge is already in the wires. What it supplies is energy, and the measure of how much energy it gives each coulomb is its electromotive force or e.m.f., also measured in volts.

Potential difference is the same quantity measured the other way round: the energy each coulomb gives up in a component. A 12 V battery hands 12 joules to every coulomb; a 12 V lamp takes 12 joules back from every coulomb passing through it.

Because p.d. is a difference between two points, it is always measured across a component, never through it. A voltmeter therefore goes in parallel with whatever you are measuring, and must have a very high resistance so that it draws almost no current itself.

V = E / QE = Q Ve.m.f. is energy given to each coulomb; p.d. is energy taken from each coulomb
V
potential differenceV
E
energy transferredJ
Q
chargeC
Worked example 14 marks

A lamp is connected to a 6.0 V supply and draws 0.50 A for 2.0 minutes. Calculate the charge that flows and the energy transferred by the lamp.

  1. Convert time: 2.0 × 60 = 120 s.Amperes are coulombs per second.
  2. Q = I t = 0.50 × 120 = 60 C.
  3. E = Q V = 60 × 6.0.Each coulomb gives up 6.0 J in the lamp.
  4. E = 360 J.Equivalently E = VIt, which is the same calculation.

Q = 60 C, E = 360 J

04

Resistance and Ohm's law

Resistance measures how strongly a component opposes the flow of charge, and it is defined by R = V/I. That definition applies to every component without exception, whether or not its resistance stays constant.

Ohm's law is a narrower claim: for a metallic conductor at constant temperature, the current is directly proportional to the potential difference. The words "at constant temperature" are not decoration — leave them out and the statement is false, and the mark is lost.

The reason for the condition is that resistance rises with temperature in a metal. The metal ions vibrate more strongly when hot, so the drifting electrons collide with them more often, and each collision impedes the flow.

ComponentI–V graphWhy
Fixed resistorstraight line through the originresistance constant — obeys Ohm's law
Filament lampcurve bending towards the V axisfilament heats up, so resistance rises
Thermistorcurve bending towards the I axisit warms up, so resistance falls
Diodenothing, then a sharp rise one way onlyconducts in one direction only
V = I RR = V / IR = ρL / AR = V/I always; Ohm's law adds that R stays constant at fixed temperature
R
resistanceΩ
ρ
resistivityΩ m
L
lengthm
A
cross-sectional area

The four graphs you are asked to sketch, on one pair of axes. The orange line from the origin to the moving point is the thing to watch: R = V/I is the gradient of THAT line, not of the curve. On the fixed resistor it lies along the curve and never moves, which is what Ohm’s law looks like. On the lamp it tilts over as the filament heats and the resistance rises; on the thermistor it swings the other way as the resistance falls. On the diode there is nothing at all until about 0.7 V, and nothing ever in reverse.

Longer and thinner means more resistance

Resistance is proportional to length and inversely proportional to cross-sectional area. A wire twice as long has twice the resistance; a wire twice as thick has half. Think of water in a pipe — a long narrow pipe resists flow more than a short fat one.

05

Measuring in a circuit

Two instruments, two rules, and getting them the wrong way round is one of the most reliably penalised errors in the subject.

An ammeter measures current, so it must have the current flowing through it. It goes in series, and it must have very low resistance — otherwise it would reduce the very current it is supposed to be measuring.

A voltmeter measures potential difference, which is a difference between two points. It goes in parallel, connected across the component, and it must have very high resistance so that hardly any current is diverted through it.

When drawing a circuit diagram, use the standard symbols and keep the wires as straight lines with right-angled corners. A neat diagram is quicker to mark and quicker to check.

Key points

  1. Only electrons move; an object goes positive by losing them.
  2. Current is charge per second and is not used up in a circuit.
  3. P.d. is energy per coulomb, always measured across a component.
  4. R = V/I defines resistance; Ohm's law adds "at constant temperature".
  5. Ammeter in series with low resistance; voltmeter in parallel with high resistance.

Practice questions

6 questions · 26 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Define electric current and state its unit.
Model answer

The rate of flow of electric charge, I = Q/t. Its unit is the ampere (A), where one ampere is one coulomb per second.

Examiner tip. The equation earns the definition mark and the unit earns the second.

SQ2[2 marks]
State Ohm's law and the condition under which it holds.
Model answer

The current through a metallic conductor is directly proportional to the potential difference across it, provided the temperature stays constant.

Examiner tip. Omit "at constant temperature" and you lose a mark, every time.

SQ3[2 marks]
Explain why the resistance of a metal wire increases as it gets hotter.
Model answer

The metal ions vibrate more strongly, so the moving electrons collide with them more often. Each collision impedes the flow, so the resistance rises.

Examiner tip. Name the ions and the collisions. "The atoms move more" alone is too vague to score.

Solved numericals

1 · 5 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[5 marks]
A wire of length 2.0 m and cross-sectional area 0.50 mm² has a resistance of 0.068 Ω. Calculate the resistivity of the metal, and state the resistance of a 4.0 m length of the same wire.
Full working
  1. Converts the area: 0.50 mm² = 0.50 × 10⁻⁶ m²the step most often dropped[1]
  2. Uses ρ = RA/L[1]
  3. ρ = (0.068 × 0.50 × 10⁻⁶) / 2.0[1]
  4. ρ = 1.7 × 10⁻⁸ Ω mcopper[1]
  5. Doubling the length doubles the resistance: 0.136 Ω[1]

ρ = 1.7 × 10⁻⁸ Ω m; R = 0.14 Ω

Examiner tip. A square millimetre is 10⁻⁶ m², not 10⁻³. Getting that wrong makes the resistivity a thousand times too large.

Long questions

1 · 8 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[8 marks]
A student connects a 6.0 V battery to a filament lamp and records the current for a range of potential differences.
  1. Sketch and describe the shape of the I–V graph obtained. [3]
  2. Explain the shape in terms of what happens inside the filament. [3]
  3. At 6.0 V the current is 0.50 A. Calculate the resistance and the power at that point. [2]
Mark scheme
  1. The graph passes through the origin[1]
  2. It is a straight line at low potential difference[1]
  3. It then curves towards the V axis, so the gradient falls[1]
  4. A larger current heats the filament[1]
  5. The ions vibrate more and the electrons collide with them more often[1]
  6. So the resistance increases and the current no longer rises in proportion[1]
  7. R = V/I = 6.0 / 0.50 = 12 Ω[1]
  8. P = VI = 6.0 × 0.50 = 3.0 W[1]

(c) 12 Ω and 3.0 W

Examiner tip. The curve bends towards the V axis, not the I axis. Sketch it wrong and you contradict your own explanation in part (b).

Exam questions

1 · 7 marks

Multi-part questions with a full mark scheme.

Q1[7 marks]
A 9.0 V supply is connected in series with a 20 Ω resistor and a thermistor. At room temperature the thermistor has a resistance of 25 Ω.
  1. Calculate the current in the circuit at room temperature. [3]
  2. Calculate the potential difference across the thermistor. [2]
  3. State and explain what happens to that potential difference as the thermistor is warmed. [2]
Mark scheme
  1. Total resistance = 20 + 25 = 45 Ω[1]
  2. Uses I = V/R[1]
  3. I = 9.0 / 45 = 0.20 A[1]
  4. Uses V = IR for the thermistor[1]
  5. V = 0.20 × 25 = 5.0 V[1]
  6. The potential difference across the thermistor decreases[1]
  7. Its resistance falls as it warms, so it takes a smaller share of the supply voltage[1]

(a) 0.20 A (b) 5.0 V (c) it falls

Examiner tip. A thermistor's resistance falls when heated — the opposite of a plain metal wire. That reversal is what the question is testing.