The big idea: Resistors in series ADD: R = R₁ + R₂. Resistors in parallel give less than either: 1/R = 1/R₁ + 1/R₂.
Capacitors are the opposite way round. In parallel they ADD: C = C₁ + C₂. In series they give less than either: 1/C = 1/C₁ + 1/C₂.
The same two values — 100 and 200 — in all four arrangements, with the arithmetic worked each time.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
| Arrangement | Formula | 100 and 200 give |
|---|---|---|
| Resistors in series | R = R₁ + R₂ | 300 Ω — larger than either |
| Resistors in parallel | 1/R = 1/R₁ + 1/R₂ | About 67 Ω — smaller than the smallest |
| Capacitors in parallel | C = C₁ + C₂ | 300 µF — larger than either |
| Capacitors in series | 1/C = 1/C₁ + 1/C₂ | About 67 µF — smaller than the smallest |
The check that catches an inverted answer: A parallel resistance is always smaller than the smallest resistor in it. A series capacitance is always smaller than the smallest capacitor.
If your answer is larger than both values, you have used the wrong rule — and that check takes two seconds.
Feeling unprepared for exams?
Get a clear study plan, practice with real questions, and know exactly where you stand before exam day. No more guessing.
| Arrangement | What it is for |
|---|---|
| Two resistors in series | A voltage divider: the junction voltage is a fixed fraction of the supply, which is how a sensor signal and a reference are made |
| Resistors in parallel | Sharing current and therefore heat between two parts, and making a value you do not have in the drawer |
| Capacitors in parallel | More stored charge, which is how a power supply smoothing bank is built — several capacitors side by side |
| Capacitors in series | A higher working voltage than either could survive alone, since the supply voltage is shared between them |
The divider equation: For two resistors in series across a supply, the voltage at the junction is
Vout = Vsupply × R₂ ÷ (R₁ + R₂), where R₂ is the lower resistor.
With 9 V across 10 kΩ and 5 kΩ, the junction sits at 9 × 5 ÷ 15 = 3 V — which is the whole of how a sensor signal is produced.
How this is tested — calculating resistance and capacitance in series and parallel. It comes up two ways:
Paper 1 — multiple choice
- Calculate a series or parallel total.
- Find the voltage at a divider junction.
Paper 2 — analysing a product
- Calculate a total and say what the arrangement is for.
- Design a divider to give a stated output voltage.
The trap: Applying the resistor rule to capacitors. They behave the opposite way round, and it is the single commonest error in this statement.
A 5 V supply must produce a 1.5 V reference for a comparator, using resistors in the tens of kilohms. Apply the divider equation to choose a pair, and explain why the values cannot be arbitrarily large or small.
Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.