This mini-lesson covers AQA 3.5 — Electricity: current, charge and potential difference, resistance and resistivity, the I–V characteristics of ohmic conductors, filament lamps and diodes, series and parallel circuits, emf and internal resistance, and potential dividers.
Section 3.5 is compulsory and appears on Paper 1.
Where this sits: AQA sections 3.1–3.8 are compulsory for every A-level Physics student. (Sections 3.9–3.13 are the five options — you study just one of those.)
Work through each screen, answer the questions as you go (many are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
3.5.1 · basics
Current, charge, potential difference and resistance
Current is the rate of flow of charge. Potential difference is the energy transferred per unit charge between two points.
I = ΔQ ÷ Δt · V = W ÷ Q · R = V ÷ I1 A = 1 C s−1 · 1 V = 1 J C−1 · 1 Ω = 1 V A−1
Charge is quantised: every charge is a whole-number multiple of e = 1.60×10−19 C.
Power in a component: P = VI = I²R = V²/R; energy transferred W = VIt.
Ohm’s law is not a universal truth — it is the special case where I ∝ V, which holds for a metal at constant temperature.
Worked example
A current of 0.50 A flows for 2.0 minutes through a 9.0 V lamp.
Q = It = 0.50 × 120 = 60 C. Energy W = VQ = 9.0 × 60 = 540 J.
Calculate
Your turn — calculation 1
1A current of 0.50 A flows for 2.0 minutes through a component with a pd of 9.0 V across it. Calculate the energy transferred.
J
Hint: Q = It = 0.50 × 120 s = 60 C. Then W = VQ.
3.5.1 · resistivity
Resistivity and superconductivity
R = ρL ÷ A → ρ = RA ÷ Lρ = resistivity, unit Ω m — a property of the material, not of the shape
Resistance rises with length and falls with cross-sectional area (a fat wire is an easier path).
For a metal, heating increases lattice vibrations, so electrons collide more often → resistivity rises.
For a thermistor (NTC) or semiconductor, heating releases many more charge carriers → resistance falls sharply. This dominates over the extra collisions.
Superconductivity: below a critical temperature the resistivity drops abruptly to zero. Uses: loss-free power cables, very strong electromagnets (MRI scanners, particle accelerators).
Worked example
Copper wire, ρ = 1.70×10−8 Ω m, L = 2.0 m, d = 0.40 mm.
A = π(0.20×10−3)² = 1.26×10−7 m².
R = (1.70×10−8 × 2.0) ÷ 1.26×10−7 = 0.27 Ω.
Calculate
Your turn — calculation 2
2A copper wire has resistivity 1.70 × 10−8 Ω m, length 2.0 m and diameter 0.40 mm. Calculate its resistance, to 2 significant figures.
Ω
Hint: A = πr² = π(0.20×10−3)² = 1.26×10−7 m². Then R = ρL ÷ A.
3.5.1 · I–V characteristics
I–V characteristics
Plotting current against pd fingerprints a component:
Ohmic conductor (metal wire at constant T): a straight line through the origin. Gradient = 1/R, and R is constant.
Filament lamp: an S-shaped curve that flattens as current rises. The filament heats up, so its resistance increases.
Semiconductor diode: conducts only in forward bias, and only above a threshold pd of about 0.6 V. In reverse bias the resistance is enormous and virtually no current flows.
Circuit skill: resistance at any point on an I–V graph is V ÷ I for that point — not the gradient of the curve, unless the line is straight through the origin. Students lose marks here every year.
Sort it
Which component does this describe?
Tap a statement, then tap the component it belongs to.
📏 Ohmic conductor
💡 Filament lamp
➡️ Semiconductor diode
Quick check
Think it through
?The temperature of an NTC thermistor is increased. What happens to its resistance, and why?
Series: the current is the same everywhere; the pds add to the supply pd.
Parallel: the pd is the same across each branch; the currents add (conservation of charge at a junction — Kirchhoff’s first law).
Adding a resistor in parallel always decreases the total resistance — you are giving charge an extra route.
Conservation of energy round any loop: the sum of the emfs = the sum of the pds (Kirchhoff’s second law).
Worked example
A 6.0 Ω and a 3.0 Ω resistor in parallel: 1/R = 1/6 + 1/3 = 3/6 → R = 2.0 Ω.
That pair is in series with a 4.0 Ω resistor: Rtotal = 2.0 + 4.0 = 6.0 Ω.
Calculate
Your turn — calculation 3
3A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel. This combination is in series with a 4.0 Ω resistor. Calculate the total resistance of the circuit.
Ω
Hint: Parallel pair: 1/R = 1/6 + 1/3 = 1/2, so R = 2.0 Ω. Then add the series resistor.
3.5.2 · emf & internal resistance
EMF and internal resistance
A real cell has internal resistance r. Some of the energy per coulomb is used inside the cell itself.
ε = I(R + r) → V = ε − Irε = emf (energy per unit charge produced) · V = terminal pd · Ir = the lost volts
Plot V against I: the y-intercept is ε and the gradient is −r. This is the standard required-practical graph.
A short circuit (R = 0) gives the maximum current I = ε/r — which is why a shorted battery gets dangerously hot.
High-current devices (car starter motors) need cells with low internal resistance.
Worked example
ε = 12 V, r = 0.50 Ω, external R = 5.5 Ω.
I = 12 ÷ (5.5 + 0.5) = 12 ÷ 6.0 = 2.0 A. Terminal pd V = 12 − (2.0 × 0.50) = 11 V.
Calculate
Your turn — calculation 4
4A cell of emf 12 V and internal resistance 0.50 Ω is connected to an external resistor of 5.5 Ω. Calculate the terminal potential difference.
V
Hint: I = ε ÷ (R + r) = 12 ÷ 6.0 = 2.0 A. Then V = ε − Ir.
Quick check
Think it through
?A student plots terminal pd V (y-axis) against current I (x-axis) for a cell. What do the intercept and gradient give?
3.5.2 · potential dividers
Potential dividers
Two resistors in series share the supply pd in proportion to their resistances.
Vout = Vin × R2 ÷ (R1 + R2)Vout is the pd across R2
Replace one resistor with an LDR or a thermistor and Vout becomes a sensor output that changes with light or temperature.
A potentiometer (a sliding contact on a resistive track) gives a continuously variable Vout from 0 up to Vin.
Careful: as more light falls on an LDR its resistance falls, so the pd across the LDR falls (and the pd across the fixed resistor rises).
Worked example
12 V across a 4.0 kΩ and an 8.0 kΩ resistor in series. Pd across the 8.0 kΩ:
5A 12 V supply is connected across a 4.0 kΩ resistor in series with an 8.0 kΩ resistor. Calculate the potential difference across the 8.0 kΩ resistor.
V
Hint: Vout = 12 × 8.0 ÷ (4.0 + 8.0).
Match it
Match each definition to its quantity
Tap a card on the left, then its partner on the right.
Statement
Answer
Quick check
Think it through
?Below its critical temperature, a superconductor has zero resistivity. Which application does this make possible?
Quick check
Think it through
?In a potential divider, an LDR (light-dependent resistor) is in series with a fixed resistor, and Vout is taken across the LDR. What happens to Vout when the light intensity increases?
Recap
The big ideas to know
Basics: I = ΔQ/Δt · V = W/Q · R = V/I · P = VI = I²R = V²/R
Resistivity: R = ρL/A · metal: R rises with T · thermistor/semiconductor: R falls with T
I–V: ohmic = straight line · lamp = flattening curve · diode = forward bias only, ~0.6 V
Circuits: series: current same, pds add · parallel: pd same, currents add, total R falls