The big idea: Four rules cover resistors and capacitors in series and parallel. Capacitors are the opposite way round to resistors.
Resistors
- In series they ADD: R = R₁ + R₂.
- In parallel they give less than either: 1/R = 1/R₁ + 1/R₂.
Capacitors
- 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.
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| 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.
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| 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.
Worked example: With 9 V across 10 kΩ and 5 kΩ, the junction sits at 9 × 5 ÷ 15 = 3 V. That 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.
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