Electricity Revision Notes — 9702 Paper 1

Electric current and Q = It

Syllabus requirement · §9.1

“

understand that an electric current is a flow of charge carriers; understand that the charge on charge carriers is quantised; recall and use Q = It

”

Charge on the move

Press a switch and a lamp lights instantly — something is moving through the wires. That something is charge, carried by particles the syllabus calls charge carriers. In a metal wire the carriers are free electrons, tiny charged particles that drift through the lattice of fixed ions; in a car battery's acid the carriers are positive and negative ions moving through the liquid; in a neon sign they are ions and electrons racing through the gas. The carrier changes from situation to situation — but the bookkeeping never does:

Electric current is the rate of flow of charge: how much charge passes a point in the circuit each second.

Because current is charge per second, its unit is built from the coulomb (charge) and the second (time):

1\ \text{ampere} = 1\ \text{coulomb per second} \qquad (1\ \text{A} = 1\ \text{C s}^{-1})

So a current of $3.0\ \text{A}$ means $3.0\ \text{C}$ of charge flows past any point of the circuit every second. One coulomb is an enormous number of electrons — about $6$ billion billion — which is why currents of a few amperes feel ordinary while we count their carriers in powers of ten.

Q = It

charge transferred (C) = current (A) × time (s)

The definition rearranged: rate × time gives the total. Every question starts by converting I to amperes and t to seconds.

A clean demo before the exam version

A phone charger delivers $250\ \text{mA}$ for $40$ minutes. How much charge has moved through the phone?

Step 1 — convert both quantities to SI units. The formula demands amperes and seconds, and neither quantity arrives that way:

I = 250\ \text{mA} = 250 \times 10^{-3}\ \text{A} = 0.250\ \text{A}

t = 40\ \text{min} = 40 \times 60 = 2400\ \text{s}

Step 2 — substitute into $Q = It$ :

Q = 0.250 \times 2400 = 600\ \text{C}

Two conversions, one multiplication. Most lost marks on $Q = It$ come from skipping Step 1: feeding in $250$ and $40$ gives $10\,000\ \text{C}$ , wrong by factors of $1000$ and $60$ at once.

One direction note before moving on: the electrons in a metal drift towards the positive terminal of a battery, but by long-standing convention current is drawn flowing from + around the circuit to −, as if positive charge moved. This "conventional current" direction is what every circuit diagram and every mark scheme uses. The calculations in this note work the same either way, because only the size of the flow enters $Q = It$ .

Charge comes in packets

Charge is not a continuous fluid. It always belongs to a particle — an electron, a proton, an ion — and those particles carry fixed, unchangeable amounts:

Charge is quantised: the charge on any object is an integer multiple of the elementary charge $e = 1.60 \times 10^{-19}\ \text{C}$ — the magnitude of the charge on one electron ( $-e$ ) or one proton ( $+e$ ).

An ion that has lost two electrons carries exactly $+2e$ ; a dust grain with a thousand spare electrons carries exactly $-1000e$ . There is no particle anywhere carrying $0.5e$ , and so there is no possible carrier charge of

1.5 \times e = 2.4 \times 10^{-19}\ \text{C}.

That last sentence is the entire content of a recurring Paper 1 question: divide the given charge by $e$ and demand a whole number.

\frac{-4.8 \times 10^{-19}}{-1.6 \times 10^{-19}} = +3 \quad \checkmark \text{ allowed (three extra electrons)}

\frac{-2.4 \times 10^{-19}}{-1.6 \times 10^{-19}} = +1.5 \quad \boldsymbol{\times}\;\text{impossible — no such carrier}

The same arithmetic runs forwards when you want a count of carriers rather than a test of one. If total charge $Q$ has flowed past a point, the number of electrons is

N = \frac{Q}{e}

— divide by $e$ , which flips the power of ten up by $+19$ . For the phone charger above,

N = \frac{600}{1.60 \times 10^{-19}} = 3.75 \times 10^{21}\ \text{electrons}

passed through the phone in those 40 minutes. Students who reach for $600 \times 1.6 \times 10^{-19}$ instead get about $10^{-16}$ "electrons": less than one particle, which is impossible. One coulomb is about $6 \times 10^{18}$ electrons, so any everyday charge contains a huge number of them.

Left: current as a flow of charge carriers — free electrons drifting through a metal wire past a marked point; the arrow labelled I shows the conventional current direction. Right: charge is quantised — dividing a charge by e must give a whole number of carriers, so −2.4×10⁻¹⁹ C cannot be the charge on any particle.

Charge through a wire in five minutes

9702/23 O/N 2024 Q6(b)(ii)2 marks

A cylindrical metal wire of length $2.4\ \text{m}$ and cross-sectional area $8.0 \times 10^{-6}\ \text{m}^2$ has a resistance of $0.33\ \Omega$ . There is a current in the wire of $4.7\ \text{A}$ .

Calculate the charge that passes through the wire in a time of $5.0$ minutes.

Show full working

1
Convert the time to seconds first.
$t = 5.0\ \text{min} = 5.0 \times 60 = 300\ \text{s}$
The current is already in amperes, so only the time needs converting. Doing this before touching the formula is the habit that keeps the minutes-slip out of your working.
2
State the rule and substitute. $Q = It$ with $I = 4.7\ \text{A}$ and $t = 300\ \text{s}$ :
$Q = 4.7 \times 300$
C1 is for writing Q = It (or this substituted line). The method mark is earned here, before the final number appears.
3
Evaluate:
$Q = 1410\ \text{C} \approx 1400\ \text{C} \ (= 1.4 \times 10^3\ \text{C}) \quad \text{(A1)}$
The data (4.7 A, 5.0 min) are given to 2 significant figures, so give the answer to 2 s.f.: 1400 C, the mark-scheme value.

Answer

$Q = 4.7 \times 300 = 1400\ \text{C}$ (to 2 s.f.).

Every Q = It question is the same three moves: convert I to A, convert t to s, multiply. Multiple-choice distractors are often the values you get with an unconverted time or current.

Common mistakes

Substituting minutes or hours straight into $Q = It$ : " $Q = 4.7 \times 5.0 = 23.5$ C".
Convert first: $5.0$ min $= 300$ s, so $Q = 4.7 \times 300 = 1400$ C.
Times in minutes and hours appear in many Q = It questions on both papers, precisely to test this. Seconds only, always.
Leaving a prefix inside the substitution: " $Q = 250 \times 2400$ " for $250$ mA.
$250\ \text{mA} = 0.250\ \text{A}$ before substituting: $Q = 0.250 \times 2400 = 600$ C.
mA → ×10⁻³, kA → ×10³. Write the converted value down explicitly; mental prefix handling mid-substitution is where the factor of 1000 escapes.
Multiplying by $e$ when asked for the NUMBER of electrons: $N = Q \times e$ .
Divide: $N = \dfrac{Q}{e}$ . Charge per electron is $e$ , so the count is $Q/e$ .
Check the size: one coulomb is about 6×10¹⁸ electrons. An answer like 600 × 1.6×10⁻¹⁹ ≈ 10⁻¹⁶ 'electrons' is impossible — you cannot have a fraction of a particle.
Accepting any small value as a possible carrier charge.
Test for quantisation: divide by $1.6 \times 10^{-19}$ C. Only WHOLE-number multiples can exist.
The M/J 2025 MCQ hinges entirely on this: −2.4×10⁻¹⁹ C is 1.5 elementary charges — impossible. The check takes five seconds; do it for every option.

Your turn

A unit-definition recall, two conversion drills, and the quantisation test.

19702/12 O/N 2021 Q331 mark
What is a description of the coulomb?

Options

A the electric charge of one electron
B the electric charge transferred by a current of one ampere in one second
C the kinetic energy gained by an electron accelerated through a potential difference of one volt
D the kinetic energy of an electron moving at a speed of one metre per second
Stuck? Show hint
Rearrange Q = It for Q, then read off what 1 A for 1 s means.
Show solution
1. 1
  From $Q = It$ : with $I = 1\ \text{A}$ and $t = 1\ \text{s}$ , the charge transferred is $Q = 1\ \text{C}$ . So the coulomb is the charge transferred by a current of one ampere in one second — B.
  Option A fails because one electron carries only 1.6×10⁻¹⁹ C — a coulomb is ~6×10¹⁸ electrons, not one. Options C and D describe energies, not charge (C is the electronvolt, an energy unit you meet at A Level).
Answer
B — the electric charge transferred by a current of one ampere in one second.
29702/14 O/N 2025 Q351 mark
A wire carries a current of $5.6\ \text{A}$ . What is the number of conduction electrons that pass a point on the wire in a time of $20\ \text{s}$ ?

Options

A $\phantom{1}1.8 \times 10^{18}$ B $2.2 \times 10^{19}$ C $3.5 \times 10^{19}$ D $7.0 \times 10^{20}$
Stuck? Show hint
First Q = It for the total charge; then N = Q/e. Keep both powers of ten under control.
Show solution
1. 1
  Total charge: $I = 5.6\ \text{A}$ and $t = 20\ \text{s}$ are already in SI units, so
  $Q = It = 5.6 \times 20 = 112\ \text{C}$
  No conversion needed here, so the difficulty is in the next step.
2. 2
  Number of electrons: divide the charge by the charge of ONE electron:
  $N = \frac{Q}{e} = \frac{112}{1.6 \times 10^{-19}} = 7.0 \times 10^{20} \quad \Rightarrow \quad \textbf{D}$
  Dividing by 1.6×10⁻¹⁹ pushes the power of ten UP by 19: 112/1.6 = 70, so 70×10¹⁹ = 7.0×10²⁰. Option C (3.5×10¹⁹) is 5.6/(1.6×10⁻¹⁹): the time was left out.
Answer
D — $N = \frac{112}{1.6 \times 10^{-19}} = 7.0 \times 10^{20}$ electrons.
39702/13 M/J 2025 Q331 mark
What cannot be the charge on a charge carrier?

Options

A $\phantom{1}-4.8 \times 10^{-19}\ \text{C}$
B $-3.2 \times 10^{-19}\ \text{C}$
C $-2.4 \times 10^{-19}\ \text{C}$
D $+3.2 \times 10^{-19}\ \text{C}$
Stuck? Show hint
Divide each option by 1.6×10⁻¹⁹ C. Three give whole numbers; one does not.
Show solution
1. 1
  Test every option against quantisation — divide each charge by $e = 1.6 \times 10^{-19}\ \text{C}$ :
  $\frac{4.8}{1.6} = 3 \;\checkmark \qquad \frac{3.2}{1.6} = 2 \;\checkmark \qquad \frac{2.4}{1.6} = 1.5 \;\boldsymbol{\times} \qquad \frac{3.2}{1.6} = 2 \;\checkmark$
  Sign is irrelevant to the test — carriers come in both signs, and quantisation constrains only the SIZE. Options A, B and D are integer multiples of e (three electrons, two electrons, two protons' worth).
2. 2
  The failing option is C: $2.4 \times 10^{-19}\ \text{C}$ is $1.5e$ , and no particle carries one and a half elementary charges.
  All four values look similar in size, so judging by eye does not work. Do the division for every option.
Answer
C — $-2.4 \times 10^{-19}\ \text{C} = -1.5e$ is not an integer multiple of the elementary charge.
4
(i) A laptop charger supplies a current of $1.8\ \text{A}$ while charging for $25$ minutes. Calculate the charge delivered.
(ii) The charge delivered to a different device is $450\ \text{C}$ over a time of $3.0$ minutes. Calculate the average current.
Stuck? Show hint
(i) Convert minutes to seconds first. (ii) Rearrange Q = It for I — with t in seconds.
Show solution
1. 1
  (i) Convert the time to seconds:
  $t = 25 \times 60 = 1500\ \text{s}$
  The current is already in amperes; only the time needs converting.
2. 2
  Substitute $I = 1.8\ \text{A}$ and $t = 1500\ \text{s}$ into $Q = It$ :
  $Q = 1.8 \times 1500 = 2700\ \text{C} = 2.7 \times 10^3\ \text{C}$
  2700 C sounds large but is ordinary for a charger running for 25 minutes.
3. 3
  (ii) Convert the time to seconds:
  $t = 3.0 \times 60 = 180\ \text{s}$
  Convert before rearranging, so the seconds are ready when you substitute.
4. 4
  Rearrange $Q = It$ for the current by dividing both sides by $t$ :
  $I = \frac{Q}{t}$
  The same formula, used backwards.
5. 5
  Substitute $Q = 450\ \text{C}$ and $t = 180\ \text{s}$ :
  $I = \frac{450}{180} = 2.5\ \text{A}$
  Using 3.0 (minutes) instead of 180 would give 150 A, an absurd current for a small device.
Answer
(i) $Q = 2.7 \times 10^3$ C. (ii) $I = 2.5$ A.

Paper 1 · MCQReal past-paper questions · Electric current as flow of charge; Q = It Paper 2 · TheoryReal past-paper questions · Electric current as flow of charge; Q = It

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Can you do all of these?

Define current as the rate of flow of charge, and use Q = It only after converting time to seconds and current to amperes
Test charge values for quantisation: divide by 1.6×10⁻¹⁹ C — a whole number is allowed, anything else cannot be a carrier charge
Count electrons as N = Q/e, remembering that dividing by 10⁻¹⁹ makes the power of ten 19 bigger
Derive I = Anvq: carriers in length L are nAL, their charge nALq passes a plane in time L/v
Keep I the same everywhere in a series loop; the drift speed adjusts — narrower conductor, faster drift
Convert diameter to area before using I = Anvq or R = ρL/A: halving d quarters A (A ∝ d²)
Define p.d. in the mark scheme's words: energy transferred per unit charge; use V = W/Q
Choose the power form by what you know: I and R → I²R; V and R → V²/R; V and I → VI
Compute efficiency as useful output ÷ total input (×100%), with the electrical input found from VI
Define resistance as potential difference per unit current, and the ohm as a volt per ampere
State Ohm's law with its condition: current proportional to p.d. provided the temperature is constant
Sketch the three I–V curves: straight line through the origin (metal); curve flattening in the first and third quadrants (lamp); zero, then a steep rise after a threshold in the first quadrant only (diode)
Find resistance from any I–V graph as V/I at the point (the line from the origin) — never from the tangent gradient
Explain the lamp curve as a chain: current increases → temperature increases → lattice vibrations increase → resistance increases → V/I increases
Use R = ρL/A with ρ in Ω m and A in m² — convert mm² → ×10⁻⁶ and diameters through A = πd²/4
Handle stretched-wire questions by conserving volume: doubling L halves A, so R quadruples (R ∝ L²)
State both sensor rules: LDR resistance decreases as light intensity increases; NTC thermistor resistance decreases as temperature increases
Reason sensing circuits as a chain: condition changes → sensor resistance → total resistance → current → p.d.s across each component