A-Level · 25 September 2026 · 6 min read
AS Physics circuits and resistivity: work from the model to the graph
Circuit questions become more manageable when you track three separate levels: the behaviour of a component, the connections in the circuit, and the quantities measured by the instruments. Problems arise when a rule from one level is applied automatically to another.
The examples here develop International AS Physics electricity skills through calculation and interpretation. They are original practice problems rather than extracts from exam papers. Keep the circuit assumptions visible as you work, especially temperature, meter loading and whether a source has internal resistance.
Build a resistivity calculation in two stages
A uniform wire is 1.20 m long and 0.40 mm in diameter. A potential difference of 0.72 V produces a current of 0.30 A. First calculate resistance: R = 0.72/0.30 = 2.4 Ω. Convert the diameter to 4.0 × 10⁻⁴ m and find A = πd²/4 = 1.26 × 10⁻⁷ m². Then ρ = RA/L = 2.51 × 10⁻⁷ Ω m.
The final value is approximately 2.5 × 10⁻⁷ Ω m to two significant figures. Do not use the wire's surface area or leave its diameter in millimetres. Resistance describes this piece of wire; resistivity characterises its material under the specified physical conditions. A longer sample of the same material can have a different resistance without a different resistivity.
Control heating rather than assuming it away
Using R = ρL/A to compare wire lengths assumes resistivity remains constant. In a practical investigation, current can heat a metal wire and change its resistivity. Use a suitably low current, take readings promptly and switch off between readings where appropriate. A repeated value at the same length can help reveal whether the experimental conditions are drifting.
Measure diameter at several positions and orientations with an appropriate instrument, checking any zero error. Because area depends on diameter squared, a small percentage uncertainty in diameter contributes approximately twice that percentage uncertainty to area. Measuring length carefully cannot compensate for a poorly known cross-sectional area; your improvement should address the measurement that most limits the result.
Interpret an I–V graph without reversing the gradient
For an ohmic conductor at constant temperature, a V-against-I graph is a straight line through the origin and its gradient is resistance. Swap the axes, and the I-against-V gradient becomes 1/R. For example, ΔV = 1.2 V and ΔI = 0.40 A give R = 3.0 Ω. The reciprocal gradient on the swapped graph is 0.333 A V⁻¹.
For a non-ohmic component, V/I at a chosen operating point gives its resistance there. A tangent gradient on a curved graph describes a small-change relationship and should not automatically replace that ratio. A filament lamp's changing temperature helps explain its curved characteristic; simply saying 'resistance is the gradient' without specifying axes and behaviour is incomplete.
Read internal resistance from the correct equation
For a source with emf ε and internal resistance r supplying current I, terminal potential difference is V = ε − Ir. On a V-against-I graph, the vertical intercept is ε and the gradient is −r. If the best-fit line falls from 1.50 V at zero current to 1.20 V at 0.60 A, r = 0.30/0.60 = 0.50 Ω.
The minus sign expresses a fall in terminal voltage as current increases; internal resistance itself is positive in this model. At 0.60 A, the internal energy-transfer rate is I²r = 0.18 W. Distinguish emf, energy supplied per unit charge by the source, from the terminal potential difference available to the external circuit.
Predict a potential divider before calculating
Two resistors of 2.0 kΩ and 3.0 kΩ are connected in series across an ideal 10 V supply. With no appreciable load across either resistor, the output across the 3.0 kΩ resistor is 10 × 3.0/(2.0 + 3.0) = 6.0 V. The larger resistor receives the larger share of the supply voltage because both carry the same current.
If a sensor replaces that lower resistor, first ask whether its resistance rises or falls as the measured condition changes. A falling lower resistance reduces the unloaded output across it. If a substantial load is attached across the output, combine that load in parallel with the lower resistor before applying the divider relationship. The original two-resistor ratio no longer describes the circuit.
Practise a comparison without unnecessary numbers
A replacement wire has the same material and temperature, twice the original length and half the original diameter. What happens to its resistance? Halving diameter quarters cross-sectional area. In R = ρL/A, the length factor is two and the area factor is one quarter, so resistance increases by a factor of eight. A statement about length alone would miss most of the change.
Now connect the replacement to the same ideal voltage source. Current becomes one eighth of its previous value, and P = V²/R also becomes one eighth. This conclusion requires fixed voltage. With a constant-current supply, P = I²R would instead increase eightfold. In any proportional question, finish by checking which electrical quantity is being held constant.
Questions, explained
Choose a question for a direct answer, then explore the explanation and supporting resources. Each answer has its own link to save or share.
Why is resistivity measured in ohm metres rather than ohms per metre?
Rearranging R = ρL/A gives ρ = RA/L. The units are therefore Ω × m²/m = Ω m. Resistance per unit length would have units Ω m⁻¹ and still depends on the wire's cross-sectional area. Keeping the area in the equation helps distinguish the material property from a property of one particular wire geometry.
Why does diameter uncertainty matter so much in a resistivity experiment?
The wire area is proportional to diameter squared, so its percentage uncertainty is approximately twice the diameter's percentage uncertainty for small uncertainties. That contribution then enters the resistivity calculation alongside uncertainties in resistance and length. Several diameter readings at different positions can reveal variation, but repeated readings alone do not remove an uncorrected zero error.
Does a larger resistance always mean a larger power?
No: you must specify what is fixed. At constant current, P = I²R increases with resistance. At constant potential difference, P = V²/R decreases as resistance rises. Both formulas are consistent with P = VI. In a real circuit, changing resistance can change both current and terminal voltage, so first work out how the source and other components behave.