AS Physics: Electricity — Revision Notes

Condensed for the final weeks. For the full explanation, use the Electricity study guide.

Definitions and equations

current            I = Q / t          (rate of flow of charge)
                   I = n A v q        (n = charge carrier density,
                                       v = drift velocity)
potential diff.    V = W / Q          (work done per unit charge)
e.m.f.             E = W / Q          (energy converted per unit charge
                                       FROM other forms TO electrical)
resistance         R = V / I
resistivity        R = rho L / A     (rho = resistivity, in Ohm m)
power              P = VI = I^2 R = V^2 / R
energy             W = VIt

e.m.f. vs p.d. — e.m.f. is energy supplied to the circuit per coulomb; p.d. is energy transferred from the circuit per coulomb. The distinction is examined directly, and a source’s e.m.f. only equals the p.d. across its terminals when no current flows or when the source has no internal resistance — otherwise some energy per coulomb is lost inside the source itself. See the Electricity study guide for the full syllabus coverage and worked reasoning behind every equation and table above.

Ohm’s law and its limits

The current through a conductor is proportional to the potential difference provided physical conditions, especially temperature, remain constant. This single sentence is the entire statement examiners expect verbatim in meaning, not just in spirit.

The temperature condition is part of the statement — omitting it loses the mark, since every I-V characteristic that curves away from a straight line does so because temperature (or, for a diode, the internal conduction mechanism) is not staying constant.

I–V characteristics

Component	Graph shape	Why
Metallic conductor at constant T	Straight line through origin	Obeys Ohm’s law
Filament lamp	Curve flattening as V rises	Temperature rises → lattice ions vibrate more → more collisions → R increases, so the gradient falls as V rises
Semiconductor diode	Zero until ~0.6 V, then steep rise	Conducts in forward bias only, with negligible current in reverse bias
Thermistor (NTC)	Curve, R falls as T rises	More charge carriers released
LDR	Curve, R falls as light intensity rises	More charge carriers released by light energy

Resistivity

R = rho L / A

Resistance depends on the object’s dimensions; resistivity is a property of the material and is independent of shape. Doubling the length doubles R; doubling the diameter quarters R (area ∝ d²).

Worked example: a wire of resistivity 1.7 x 10^-8 Ohm m, length 2.0 m
and cross-sectional area 5.0 x 10^-7 m2.
R = rho L / A = (1.7 x 10^-8 x 2.0) / (5.0 x 10^-7) = 0.068 Ohm

A longer, thinner wire always has a higher resistance than a shorter, thicker one of the same material, since R is directly proportional to length and inversely proportional to area.

Charge carriers

I = n A v q

For a given current, a thinner wire (smaller A) means a higher drift velocity — the standard exam application. Metals have very large n, so drift velocity is tiny (millimetres per second) despite the near-instant effect — the electric field that starts the current flowing propagates through the conductor far faster than any individual electron actually moves.

Exam traps

• Stating Ohm’s law without the constant-temperature condition.
• Confusing e.m.f. with p.d.
• Saying a filament lamp “does not obey Ohm’s law because it is a lamp” — the reason is the temperature rise increasing resistance.
• Forgetting A ∝ d² when a diameter is doubled.
• Using diameter instead of radius in A = πr², or forgetting to halve a given diameter before squaring it.
• Resistivity has units Ω m, not Ω/m.
• Confusing an LDR with a thermistor — an LDR’s resistance responds to light intensity, a thermistor’s to temperature; both fall as the relevant quantity rises, but the trigger is different.
• Forgetting to convert a cross-sectional area given as a diameter in mm into m² before substituting into R = ρL/A.

Self-test

1. Define potential difference.
2. Explain the shape of the I–V graph for a filament lamp.
3. A wire’s diameter is doubled. What happens to its resistance?
4. State Ohm’s law in full.
5. A current is constant. What happens to drift velocity if the wire narrows?
6. A wire of resistivity 2.8 × 10⁻⁸ Ω m, length 1.5 m and cross-sectional area 2.0 × 10⁻⁷ m² is connected in a circuit. Calculate its resistance.
7. State how the resistance of an LDR changes as light intensity increases, and why.

Answers: 1. The work done (energy transferred) per unit charge as charge passes between two points, V = W/Q. 2. As current increases, the filament’s temperature rises; lattice ions vibrate with greater amplitude, so electrons collide more frequently; resistance increases and the graph curves away from the straight line. 3. Area is proportional to d², so quadrupling the area quarters the resistance. 4. Current is directly proportional to potential difference provided physical conditions, particularly temperature, remain constant. 5. It increases — from I = nAvq, if I and n are fixed and A falls, v must rise. 6. R = ρL/A = (2.8 × 10⁻⁸ × 1.5) ÷ (2.0 × 10⁻⁷) = 0.21 Ω. 7. It falls — more light energy releases more charge carriers, increasing the number density n available to carry current.