Circuits questions are some of the most predictable marks in the Combined Science Physics paper. The physics never changes: the same four relationships appear year after year, and once you can spot whether components are in series or parallel, the arithmetic is usually straightforward. The marks are lost elsewhere, in setting out, in unit slips and in misreading the circuit diagram.
Here is the method our tutors use, the one that works under time pressure in Paper 2.
The four rules you must know cold
Everything in this topic comes from these. Memorise them as a pair of contrasts, not as eight separate facts.
Series
- Current is the same at every point: I is constant.
- Potential difference divides: V_total = V1 + V2 + ...
- Total resistance: R = R1 + R2 + ...
Parallel
- Potential difference is the same across each branch: V is constant.
- Current divides: I_total = I1 + I2 + ...
- Total resistance: 1/R = 1/R1 + 1/R2 + ...
And tying them together, Ohm's Law: V = IR, plus power P = VI = I²R.
One sanity check that saves you every time: the combined resistance of a parallel arrangement is always smaller than the smallest individual resistor. If you calculate two 6 Ω resistors in parallel and get 12 Ω, you have added when you should have used the reciprocal formula.
Step-by-step method for any circuit question
- Redraw or label the diagram. Mark the EMF of the source, and write the resistance next to each component.
- Identify groupings. Ask: do these two components share both junctions? If yes, they are in parallel. Does all the current that passes through one have to pass through the other? If yes, they are in series.
- Collapse the parallel sections first into a single equivalent resistance.
- Add up the series chain to get total resistance.
- Find the total current from the supply using I = V/R with the EMF and total resistance.
- Work backwards to find the p.d. across each part, then the current in each branch.
Step 6 is where most students stall. The trick is to remember that once you know the current through a series section, V = IR gives you the p.d. across it immediately, and the p.d. across a parallel block is shared by every branch inside it.
A worked example
A 12 V battery is connected to a 4 Ω resistor in series with a parallel combination of a 6 Ω and a 3 Ω resistor. Find the current through the 6 Ω resistor.
Step 1, collapse the parallel block: 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, so R = 2 Ω.
Check: 2 Ω is smaller than 3 Ω, the smallest branch. Good.
Step 2, total resistance: R_total = 4 + 2 = 6 Ω.
Step 3, total current: I = V/R = 12/6 = 2.0 A.
Step 4, p.d. across the parallel block: The 4 Ω resistor takes V = IR = 2.0 × 4 = 8.0 V, so the parallel block has 12 − 8 = 4.0 V across it. (Or directly: V = 2.0 × 2 = 4.0 V. Same answer, and it is a useful cross-check.)
Step 5, current through the 6 Ω resistor: I = V/R = 4.0/6 = 0.67 A (2 s.f.).
Quick check: the 3 Ω branch carries 4.0/3 = 1.33 A, and 0.67 + 1.33 = 2.0 A, which matches the total current. If your branch currents do not add up to the total, something is wrong.
Common mistakes that cost real marks
- Forgetting to invert. Students compute 1/R = 0.5 and write R = 0.5 Ω. The reciprocal formula gives you 1/R, not R.
- Assuming a bigger resistor takes a bigger current. In parallel, the larger resistance takes the smaller share of the current.
- Treating the ammeter and voltmeter wrongly. An ideal ammeter has zero resistance and goes in series. An ideal voltmeter has infinite resistance and goes in parallel across the component.
- Dropping units. Write A, V, Ω or W after every answer. Bare numbers are penalised.
- Over-rounding mid-way. Keep full values in your calculator and round only at the final line, then give 2 or 3 significant figures.
Phrasing your explanation answers
Structured questions often ask why something happens, for example why a lamp gets dimmer when a second lamp is added in series. Markers want a chain of reasoning, not a one-liner. Use this pattern:
"Adding the lamp in series increases the total resistance of the circuit. Since the EMF is unchanged, from I = V/R the current decreases. Power dissipated in the lamp P = I²R therefore decreases, so the lamp is dimmer."
Three causal links, each naming a quantity and its direction of change. That structure reliably scores full marks.
For parallel branches: "Connecting the second lamp in parallel does not change the p.d. across the first lamp, so the current through it and its brightness are unchanged, but the total current drawn from the supply increases."
Practise smart
Do not just repeat questions you can already do. Take five past-paper circuit questions, cover the answers, and time yourself at roughly one and a half minutes per mark. Then check whether your working shows the formula, the substitution and the answer with units, because that is exactly how the mark scheme allocates credit.
If circuits still feel like guesswork, a tutor who sat these papers recently can spot the exact step where your reasoning breaks. Message us on WhatsApp at https://wa.link/dsgbkf to book a free trial class and work through a real circuits question together.