OCR A Level Physics A: Charge, current, drift velocity, p.d. and e.m.f. (H556) – Revision Notes

Quick recall for sections 4.1.1, 4.1.2, 4.2.1 and 4.2.2 of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. H556 is the full A level. For full explanations, read the study guide, then test yourself with the practice questions.

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Definitions

Term	Meaning
Electric current	Rate of flow of charge, I = ΔQ/Δt
Coulomb (C)	Charge delivered by a steady 1 A in 1 s, so 1 C = 1 A s
Elementary charge e	1.6 × 10⁻¹⁹ C; electron has charge −e, proton +e
Quantised charge	Net charge on any object or particle is ne, with n a whole number
Conventional current	Taken to leave the source at its positive terminal and return at its negative terminal
Electron flow	Electrons drift the opposite way, from negative towards positive
Mean drift velocity	The slow average velocity of charge carriers along a conductor due to the field
Number density n	Number of charge carriers per unit volume, in m⁻³
p.d.	Energy transferred per unit charge from electrical to other forms between two points
e.m.f.	Energy given to each unit of charge by a source, converting other forms to electrical
Volt	One joule per coulomb, J C⁻¹ (kg m² s⁻³ A⁻¹ in base units)

Equations

Equation	Use
I = ΔQ/Δt	Current from charge and time (time in seconds)
Q = ne	Charge of an object with n excess or missing electrons
Σ I in = Σ I out	Kirchhoff’s first law at a junction
I = Anev	Drift velocity, area, number density, charge
V = W/Q, so W = VQ	Energy transferred across a p.d.
E = W/Q, so W = EQ	Energy supplied by a source
eV = ½mv²	Charged particle accelerated from rest through a p.d.

Method in steps: drift velocity

1. Write I = Anev and rearrange to v = I / (A n e).
2. Convert the diameter to metres and find A = πd²/4.
3. Check n is in m⁻³ (convert from cm⁻³ by multiplying by 10⁶).
4. Substitute and give the answer in m s⁻¹.
5. For comparisons, use proportion: v ∝ I, v ∝ 1/A, v ∝ 1/n.

Method in steps: accelerated particle

1. Equate electrical energy gained to kinetic energy: qV = ½mv².
2. Use the charge q of the particle: e for an electron or proton, 2e for an alpha particle.
3. Rearrange: v = √(2qV/m).
4. Use the correct mass. For the same V an electron ends up much faster than a proton, by the square root of the mass ratio.

Key facts to recall

• Electrons are the carriers in metals. Ions are the carriers in electrolytes, and both signs of ion move, in opposite directions, so each adds to the current.
• Current in a series circuit has the same value at every point. In a parallel circuit the branch currents add to the supply current.
• Charge is conserved. Kirchhoff’s first law applies that conservation to a junction.
• A positive ion has charge +ze, where z is a whole number. An ion with charge +2e counts twice as much as a singly charged ion when you work out the current.
• Drift velocities in metals are tiny, of the order of fractions of a millimetre per second, because n is so large.
• A lamp comes on at once when the switch closes because the field travels through the circuit very quickly. The carriers already in the filament start to drift together.
• The volt measures energy per unit charge for both p.d. and e.m.f. The difference is the direction of the energy change.

Method in steps: e.m.f. and p.d. questions

1. Find the charge: Q = It, with t in seconds.
2. For energy supplied by a source use W = EQ. For energy delivered to a component use W = VQ.
3. Compare the two energies. The difference is energy transferred to thermal energy inside the source.
4. Always give the unit joule (J) for W, coulomb (C) for Q and volt (V) for E and V.

Example. A 6.0 V supply drives 0.50 A for 4.0 minutes. Q = 0.50 × 240 = 120 C. Energy supplied = 6.0 × 120 = 720 J. If the lamp has 5.5 V across it, energy delivered = 5.5 × 120 = 660 J. The remaining 60 J is dissipated inside the supply.

Method in steps: Kirchhoff’s first law

1. Mark every current on the diagram with an arrow.
2. Write the currents into the junction on one side of an equation and the currents out on the other.
3. Solve for the unknown. If your answer is negative, the current flows in the opposite direction to the arrow you drew.

Charge and electron arithmetic

• Number of electrons N = Q / e. For 2.0 μC (2.0 × 10⁻⁶ C), N = 1.25 × 10¹³.
• Charge of N electrons = −Ne.
• Electrons passing each second = I / e. For 5.0 μA this is 3.1 × 10¹³ s⁻¹.
• Current from a flow of charged particles: I = (number per second) × (charge on each).

Worked reminders

• A current of 0.60 A for 90 s moves 54 C. At 1.6 × 10⁻¹⁹ C per electron that is 3.4 × 10²⁰ electrons.
• A drift speed in a 1.0 mm² metal wire carrying 3.0 A, n = 8.5 × 10²⁸ m⁻³: v = 3.0 / (1.0 × 10⁻⁶ × 8.5 × 10²⁸ × 1.6 × 10⁻¹⁹) = 2.2 × 10⁻⁴ m s⁻¹.
• Energy of a 3.0 C charge across 9.0 V: 27 J.

Must-know distinctions

Pair	Distinction
Conventional current vs electron flow	Opposite directions in a metal; both describe the same current
Drift velocity vs thermal speed	Carriers move rapidly at random; drift is the small net movement along the wire
Metal vs electrolyte	Electrons carry current in a metal; positive and negative ions both carry it in an electrolyte
e.m.f. vs p.d.	e.m.f. is energy given to charge in a source; p.d. is energy taken from charge across a component
Conductor vs semiconductor vs insulator	Large, much smaller, almost zero n
Ammeter vs voltmeter	Ammeter in series; voltmeter in parallel

Circuit symbol checklist

Be able to draw: cell, battery, switch, fixed resistor, variable resistor, filament lamp, diode, LED, ammeter, voltmeter, thermistor, LDR and fuse. Draw junction dots, use a ruler, and keep circuits as complete loops.

Quick self-test

1. A current of 0.35 A flows for 12 minutes. How much charge passes?
2. Can an object carry a charge of 4.0 × 10⁻¹⁹ C? Explain.
3. At a junction, currents of 6.0 A and 2.5 A enter and 4.1 A leaves in one wire. What current leaves in the other wire?
4. A wire carries a steady current. It is replaced by a wire of the same metal with twice the diameter. How does the drift velocity change?
5. A 1.0 mm² metal strip (n = 8.5 × 10²⁸ m⁻³) carries 2.0 A. Calculate the drift velocity (1.0 mm² = 1.0 × 10⁻⁶ m²).
6. How much energy is transferred when 18 C passes across a p.d. of 5.0 V?
7. A source of e.m.f. 3.0 V supplies 150 J to the charge it moves. How much charge moved?
8. A proton is accelerated from rest through 400 V. Find its speed (proton mass 1.67 × 10⁻²⁷ kg).
9. Express the volt in SI base units.
10. State why a semiconductor conducts better when warmed.
11. State the kinetic energy gained by an electron accelerated through 2.0 kV, in joules.

Answers

1. Δt = 720 s, Q = 0.35 × 720 = 252 C.
2. 4.0 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 2.5, not a whole number, so no.
3. 6.0 + 2.5 − 4.1 = 4.4 A.
4. Area ×4, so v ÷ 4: the drift velocity falls to one quarter.
5. v = 2.0 / (1.0 × 10⁻⁶ × 8.5 × 10²⁸ × 1.6 × 10⁻¹⁹) = 1.5 × 10⁻⁴ m s⁻¹.
6. W = VQ = 5.0 × 18 = 90 J.
7. Q = W/E = 150 / 3.0 = 50 C.
8. v = √(2 × 1.6 × 10⁻¹⁹ × 400 / 1.67 × 10⁻²⁷) = 2.8 × 10⁵ m s⁻¹.
9. J C⁻¹ = kg m² s⁻² / (A s) = kg m² s⁻³ A⁻¹.
10. Heating releases more carriers, so n increases and, for the same drift velocity, the current rises.
11. E = eV = 1.6 × 10⁻¹⁹ × 2000 = 3.2 × 10⁻¹⁶ J.

Where marks are usually lost

• Not converting minutes or hours to seconds.
• Forgetting that e = 1.6 × 10⁻¹⁹ C must also be in the denominator of I = Anev.
• Using the diameter in place of the area.
• Saying electrons “use up” current. Current is the same either side of a component in series, and Kirchhoff’s first law holds at junctions.
• Defining e.m.f. as “the p.d. across the terminals”. Say energy per unit charge supplied to charge in the source.
• Saying drift velocity is “the speed of the electrons”. It is the mean drift along the wire.
• Writing the symbol for e.m.f. as E and then confusing it with energy in an equation.
• Quoting n for semiconductors as “zero”. It is small, not zero, and it rises with temperature.

Official syllabus

OCR Advanced GCE in Physics A (H556), Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Sections 4.1.1, 4.1.2, 4.2.1 and 4.2.2.