The big idea: Old fairy lights go dark completely when one bulb fails; the lamps in your house don't — switch one off and the rest stay lit. That's the difference between series and parallel wiring.
Two ways to join components: Series = joined in one single loop, end to end — only one path for the charge.
Parallel = joined side by side on separate branches — the charge has a choice of paths.
Series (one loop)
- Same current through every component
- The p.d. (voltage) splits between them
- Resistances add: Rs = R₁ + R₂
- Total resistance is larger than any one part
Parallel (side-by-side branches)
- Same p.d. across every branch
- The current splits between the branches
- Reciprocals add: 1/Rp = 1/R₁ + 1/R₂
- Total resistance is smaller than any one part
Series: one single loop, so the SAME current flows through R₁ and R₂. The cell's p.d. is shared — it splits between the two resistors and the two p.d.s add up to 6.0 V.
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Spot it: Series → one path → same current, voltage shares out.
Parallel → many paths → same voltage, current shares out.
('p.d.' = potential difference = voltage across a component.)
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Replace several resistors with a single equivalent resistance — one resistor that would draw the same current from the cell. The rule depends on how they are wired.
In series the resistances simply add up:
- total (equivalent) resistance in series (Ω)
- the individual resistances (Ω)
In parallel you add the reciprocals (1 ÷ each resistance), then flip the answer back over to get Rp:
- total (equivalent) resistance in parallel (Ω)
- the individual resistances (Ω)
Two things to get right: 1. Parallel: after adding the 1/R terms, flip to get Rp — a common slip is to forget the final flip.
2. Adding a resistor in parallel makes the total smaller (more paths for the charge), not bigger.
| Series (one loop) | Parallel (branches) | |
|---|---|---|
| Current | same through each | splits between branches |
| P.d. (voltage) | splits between them | same across each |
| Combine R | Rs = R₁ + R₂ | 1/Rp = 1/R₁ + 1/R₂ |
| Total R vs the parts | bigger than any one | smaller than any one |
A 2.0 Ω and a 4.0 Ω resistor are connected together. Find the total (equivalent) resistance (a) in series, and (b) in parallel.
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Variable resistors and the potential divider
| Component | Its resistance changes with | Direction |
|---|---|---|
| Thermistor | temperature | warmer → lower resistance |
| Light-dependent resistor (LDR) | light intensity | brighter → lower resistance |
| Potentiometer | the position of a sliding contact | set by hand: a variable share of the pd |
Ideal meters: An ideal ammeter has zero resistance, so there is no pd across it. An ideal voltmeter has infinite resistance, so no current goes through it. Neither changes the circuit it measures.
A 9.0 V battery (no internal resistance) drives a 120 Ω resistor in series with a parallel pair: a 300 Ω resistor and a thermistor. A voltmeter across the parallel pair reads 3.0 V. Calculate the current in the battery and the resistance of the thermistor.
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How this is tested — one of the most-tested ideas in the whole topic:
Paper 1A
- Rank combinations of identical resistors by total resistance.
- Find a branch-current ratio in a parallel network.
Paper 2
- A labelled circuit: find the equivalent resistance, an ammeter reading, or a missing p.d. / the cell's emf.
The classic trap: Forgetting to flip the parallel reciprocal, or thinking the current splits in a series circuit — it doesn't; it's the same everywhere in one loop.
The series shortcut: In a single series loop the current is the same everywhere, so an ammeter reads that one current wherever you place it.
The supply p.d. shares out between the resistors in proportion to their resistance — and the separate p.d.s add up to the supply.
Reducing a network: Work from the inside out: replace each simple series or parallel group by one resistor, redraw, repeat. Two identical resistors R in parallel give R/2; in series, 2R. Four identical R as (R + R) in parallel with R, then R in series: 2R/3 + R = 5R/3.
One loop ⇒ the ammeter reads the same current everywhere. The voltmeter reads the p.d. across R₁; the two resistor p.d.s add up to the 9.0 V supply.
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The words to know
| Word | What it means |
|---|---|
| Equivalent resistance | The one resistor that would draw the same current from the supply |
| Potential divider | Resistors in series sharing a pd in the ratio of their resistances |
| Thermistor / LDR | A resistor whose resistance falls as temperature / light intensity rises |
| Ideal voltmeter | Infinite resistance: takes no current |
A 6.0 Ω resistor (R₁) and a 3.0 Ω resistor (R₂) are in series across a 9.0 V supply (internal resistance negligible). (a) Find the current shown on the ammeter. (b) Find the p.d. the voltmeter reads across R₁.
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