Potential dividers show up in almost every Pure Physics paper, and they are one of the few topics where a single formula plus one line of clear reasoning can bag you full marks. Yet they are also where a lot of students lose easy marks, usually because they jump straight to numbers without deciding which resistor the output voltage is taken across.
Here is the method our tutors use, the same one we used when we sat 5054 ourselves.
What a potential divider actually is
Two (or more) resistors in series across a supply. Because the components are in series, the same current flows through both. Since V = IR, the resistor with the larger resistance takes the larger share of the supply voltage.
That one sentence is the whole topic. Everything else is bookkeeping.
The formula for the p.d. across resistor R₁ in a two-resistor divider is:
V₁ = [R₁ / (R₁ + R₂)] × V_supply
Read it as: the resistance you care about, over the total resistance, times the supply.
The single most common mistake
Students write R₂ on top when the output is taken across R₁. Before you touch the calculator, physically circle the component that the output terminals are connected across on the diagram. Whatever you circled goes on the numerator.
Worked example 1: a plain divider
A 12 V supply is connected in series with a 4.0 Ω resistor and an 8.0 Ω resistor. Find the p.d. across the 8.0 Ω resistor.
- Total resistance = 4.0 + 8.0 = 12 Ω
- Current = 12 / 12 = 1.0 A
- V across 8.0 Ω = 1.0 × 8.0 = 8.0 V
Or in one line: 12 × (8.0 / 12.0) = 8.0 V.
Both methods earn full marks. The current method is slower but safer when the question later asks for power or current, so if you are unsure, find the current first.
Worked example 2: an LDR sensing circuit
A 9.0 V supply is connected to a 2.0 kΩ fixed resistor in series with an LDR. The output is taken across the fixed resistor.
In darkness, the LDR has resistance 10 kΩ:
V_out = 9.0 × [2.0 / (2.0 + 10)] = 1.5 V
In bright light, the LDR has resistance 0.50 kΩ:
V_out = 9.0 × [2.0 / (2.0 + 0.50)] = 7.2 V
So the output rises as it gets brighter. This is exactly the kind of circuit used in automatic street lighting and camera light meters, and examiners love asking you to explain the change qualitatively.
The explanation questions: how to phrase it
When the question says "Explain what happens to the output voltage when the temperature increases", you are being marked on a chain of reasoning, not on a number. Write it as a chain:
For a thermistor (NTC type, which is what the syllabus assumes) in series with a fixed resistor, with output across the fixed resistor:
- Temperature increases, so the resistance of the thermistor decreases.
- The total resistance of the circuit decreases.
- Since the supply e.m.f. is constant, the current in the circuit increases.
- The p.d. across the fixed resistor, V = IR, increases (R is unchanged, I has increased).
- Therefore the output voltage increases.
If the output is taken across the thermistor instead, the answer flips: the thermistor takes a smaller share of the supply, so the output voltage decreases.
Key phrases that pick up marks:
- "resistance of the thermistor decreases as temperature increases"
- "the thermistor takes a smaller share of the supply p.d."
- "total p.d. across the two components remains equal to the e.m.f."
Avoid vague statements like "the voltage changes because of the resistance". No mark.
Three traps to watch for
1. The voltmeter is not ideal. If a question tells you the voltmeter has a stated resistance, it is in parallel with the component. Combine them first using 1/R = 1/R₁ + 1/R₂, then apply the divider formula. The measured reading will be lower than the true value.
2. Units. Mixing kΩ and Ω is the biggest source of careless errors. Either convert everything to ohms, or keep everything in kΩ consistently, since the ratio cancels the prefix anyway.
3. Rheostat as a divider. A variable resistor connected as a potentiometer (three terminals, slider) gives a continuously variable output from 0 V to the full supply. A rheostat (two terminals, in series) only limits the current and cannot give you 0 V. Know which one the diagram shows.
How to practise this properly
Do not just repeat numerical substitution. For every past-paper divider question, force yourself to also write the two-line qualitative explanation, even when it is not asked. That is the version that appears in Paper 2 structured questions, and it is where the marks are actually won.
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