Specification: AQA A-level Physics (7408) 3.5.1.1 | Paper: 1 (and 2 as assumed knowledge) | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Electric current: the rate of flow of charge, I = ΔQ ÷ Δt. A current of 1 ampere means 1 coulomb of charge passes a point each second.
Charge carriers: the charged particles that move to form a current – free (delocalised) electrons in a metal, and positive and negative ions in an electrolyte. Conventional current is from + to −, opposite to the direction the electrons move.
Quantisation of charge: charge comes in whole-number multiples of the elementary charge e = 1.60 × 10−19 C, so the number of electrons carrying a charge ΔQ is N = ΔQ ÷ e.
The equation
current = charge flow ÷ time taken I = ΔQ ÷ Δt and number of charge carriers = charge ÷ charge on each carrier N = ΔQ ÷ e
| Quantity | Symbol | Unit |
|---|---|---|
| current | I | amperes (A), equivalent to C s⁻¹ |
| charge flow | ΔQ | coulombs (C) |
| time taken | Δt | seconds (s) |
| elementary charge | e | coulombs (C), 1.60 × 10−19 C |
| number of charge carriers | N | no unit |
Rearranged: ΔQ = IΔt | Δt = ΔQ ÷ I | N = ΔQ ÷ e | Convert mA to A (× 10−3), µA to A (× 10−6) and minutes or hours to seconds before you substitute.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A charge of 36 C flows through a heater in 12 s. Calculate the current in the heater.
Show solution
F: I = ΔQ ÷ Δt
I: I = 36 ÷ 12
F: I is already the subject
A: 3.0 A
Q2 (Core) A current of 0.40 A flows through a lamp for 25 s. Calculate the charge that flows through the lamp.
Show solution
F: I = ΔQ ÷ Δt
I: 0.40 = ΔQ ÷ 25
F: ΔQ = 0.40 × 25
A: 10 C
Q3 (Core) A torch bulb draws a current of 250 mA. Calculate the time taken for 6.0 C of charge to flow through it.
Show solution
Convert first: 250 mA = 0.250 A
F: I = ΔQ ÷ Δt
I: 0.250 = 6.0 ÷ Δt
F: Δt = 6.0 ÷ 0.250
A: 24 s
Q4 (Core) A smoke detector draws a steady current of 45 µA. Calculate the number of electrons that pass a point in its circuit each second, to 2 significant figures. (e = 1.60 × 10−19 C)
Show solution
Convert first: 45 µA = 45 × 10−6 A = 4.5 × 10−5 A
Step 1 – charge in 1 s
F: I = ΔQ ÷ Δt
I: 4.5 × 10−5 = ΔQ ÷ 1.0
F: ΔQ = 4.5 × 10−5 × 1.0 = 4.5 × 10−5 C
Step 2 – number of electrons
N = ΔQ ÷ e = 4.5 × 10−5 ÷ 1.60 × 10−19 = 2.8125 × 1014
A: 2.8 × 1014 electrons each second (2 s.f.)
Watch out: divide the charge by e, do not multiply. A number of electrons has no unit, and it must be a huge number.
Q5 (Exam) A phone battery has a capacity of 3000 mA h. (a) Calculate the total charge the battery can deliver, in coulombs. (b) The phone draws an average current of 150 mA. Calculate how long a fully charged battery lasts, in hours.
Show solution
Convert first: 3000 mA = 3.000 A; 1 h = 3600 s; 150 mA = 0.150 A
Step 1 – (a) total charge
F: I = ΔQ ÷ Δt
I: 3.000 = ΔQ ÷ 3600
F: ΔQ = 3.000 × 3600
A: 10 800 C
Step 2 – (b) time
F: Δt = ΔQ ÷ I
I: Δt = 10 800 ÷ 0.150 = 72 000 s
A: 72 000 ÷ 3600 = 20 h
Watch out: mA h is a unit of charge (current × time), not of current. 1 mA h = 3.6 C.
Q6 (Exam) A student plots a graph of the charge Q that has passed through a resistor (y-axis) against time t (x-axis). The charge counter was not reset to zero at the start, so the graph is a straight line that does not pass through the origin. At t = 5.0 s, Q = 0.80 C, and at t = 20.0 s, Q = 2.60 C. Calculate the current in the resistor.
Show solution
F: I = ΔQ ÷ Δt, which is the gradient of the Q–t graph
I: I = (2.60 − 0.80) ÷ (20.0 − 5.0) = 1.80 ÷ 15.0
F: I is already the subject
A: 0.12 A
Watch out: current is the gradient of a charge–time graph, so use changes (ΔQ and Δt). Dividing a single reading, 2.60 ÷ 20.0 = 0.13 A, is wrong here because the line does not pass through the origin.
Q7 (Exam) When a motor is switched on, the current through it is a steady 2.4 A for 3.0 s. The current then falls at a steady rate to zero over the next 4.0 s. Calculate (a) the total charge that flows through the motor and (b) the number of electrons this charge represents. (e = 1.60 × 10−19 C)
Show solution
Step 1 – charge while the current is steady
F: I = ΔQ ÷ Δt
I: 2.4 = ΔQ ÷ 3.0
F: ΔQ = 2.4 × 3.0 = 7.2 C
Step 2 – charge while the current falls
the current falls steadily, so the mean current is 2.4 ÷ 2 = 1.2 A; ΔQ = 1.2 × 4.0 = 4.8 C (the area of the triangle under the I–t graph)
Step 3 – total charge and number of electrons
ΔQ = 7.2 + 4.8 = 12 C; N = ΔQ ÷ e = 12 ÷ 1.60 × 10−19
A: (a) 12 C (b) 7.5 × 1019 electrons (2 s.f.)
Watch out: charge is the area under a current–time graph. Using 2.4 A for the whole 7.0 s gives 16.8 C, which is too big.
Q8 (Exam) During the electrolysis of copper sulfate solution, 9.0 × 1019 copper ions, Cu²⁺, arrive at the cathode in 60 s. Each Cu²⁺ ion carries a charge of +2e. Assuming all the charge arriving at the cathode is carried by these ions, calculate the current, to 2 significant figures. (e = 1.60 × 10−19 C)
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Step 1 – charge on one ion
2e = 2 × 1.60 × 10−19 = 3.20 × 10−19 C
Step 2 – total charge
ΔQ = 9.0 × 1019 × 3.20 × 10−19 = 28.8 C
Step 3 – current
F: I = ΔQ ÷ Δt
I: I = 28.8 ÷ 60
F: I is already the subject
A: 0.48 A (2 s.f.)
Watch out: each Cu²⁺ ion carries 2e, not e. Forgetting the 2 gives 0.24 A, half the correct answer.
Q9 (Stretch) An electron gun produces a narrow beam of electrons. The beam current is 2.4 mA and every electron travels at 3.0 × 107 m s⁻¹. Calculate (a) the number of electrons leaving the gun each second and (b) the number of electrons in a 0.30 m length of the beam. (e = 1.60 × 10−19 C)
Show solution
Convert first: 2.4 mA = 2.4 × 10−3 A
Step 1 – (a) electrons per second
F: I = ΔQ ÷ Δt, so in 1 s, ΔQ = 2.4 × 10−3 C
I: N = ΔQ ÷ e = 2.4 × 10−3 ÷ 1.60 × 10−19
A: 1.5 × 1016 electrons per second
Step 2 – time for an electron to travel 0.30 m
Δt = distance ÷ speed = 0.30 ÷ 3.0 × 107 = 1.0 × 10−8 s
Step 3 – (b) electrons in 0.30 m of beam
every electron in that length leaves the gun within 1.0 × 10−8 s, so N = 1.5 × 1016 × 1.0 × 10−8
A: 1.5 × 108 electrons
Check with ΔQ = IΔt: ΔQ = 2.4 × 10−3 × 1.0 × 10−8 = 2.4 × 10−11 C, and 2.4 × 10−11 ÷ 1.60 × 10−19 = 1.5 × 108
Q10 (Stretch) In a chip factory, a beam of singly charged boron ions, B⁺ (charge +e), is fired into a silicon wafer. The beam current is 50 µA. (a) Calculate the number of ions hitting the wafer each second, to 2 significant figures. (b) Calculate the time needed to implant 1.0 × 1017 ions, in minutes, to 2 significant figures. (c) The beam is changed to doubly charged ions, B²⁺, with the same beam current. Calculate the new time needed to implant 1.0 × 1017 ions, in minutes, to 2 significant figures. (e = 1.60 × 10−19 C)
Show solution
Convert first: 50 µA = 50 × 10−6 A = 5.0 × 10−5 A
Step 1 – (a) ions per second
in 1 s, ΔQ = 5.0 × 10−5 C; N = ΔQ ÷ e = 5.0 × 10−5 ÷ 1.60 × 10−19 = 3.125 × 1014
A: 3.1 × 1014 ions per second (2 s.f.)
Step 2 – (b) charge needed, then time
ΔQ = Ne = 1.0 × 1017 × 1.60 × 10−19 = 0.016 C
F: I = ΔQ ÷ Δt
I: 5.0 × 10−5 = 0.016 ÷ Δt
F: Δt = 0.016 ÷ 5.0 × 10−5 = 320 s
A: 320 ÷ 60 = 5.33… minutes = 5.3 minutes (2 s.f.)
Step 3 – (c) doubly charged ions
ΔQ = N × 2e = 1.0 × 1017 × 3.20 × 10−19 = 0.032 C; Δt = 0.032 ÷ 5.0 × 10−5 = 640 s
A: 640 ÷ 60 = 10.66… minutes = 11 minutes (2 s.f.)
Watch out: the same current means the same charge per second. Each B²⁺ ion carries twice the charge, so only half as many ions arrive each second and the time doubles. It does not halve.
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