The big idea: There are two ways to connect components in a circuit.
Series = joined in one single loop, end to end — there is 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
[Diagram: phys-circuit] - Available in full study mode
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.)
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 |
Worked example — (a) two resistors in series
A 2.0 Ω and a 4.0 Ω resistor are connected in series. Find the total (equivalent) resistance.
Solution
- Start with the given series rule:
- Put in the numbers (R₁ = 2.0 Ω, R₂ = 4.0 Ω):
- Work it out — keep the unit:
Final answer
Rs = 6.0 Ω — bigger than either resistor, as series always is.
Worked example — (b) the SAME two resistors in parallel
Now connect the same 2.0 Ω and 4.0 Ω resistors in parallel instead. Find the total (equivalent) resistance.
Solution
- Start with the given parallel rule:
- Put in the numbers:
- Flip to get Rp (don't forget this step!):
Final answer
Rp ≈ 1.3 Ω — smaller than either resistor, as parallel always is.
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How this is tested: This is one of the most-tested ideas in the whole topic.
- Paper 1A: quick MCQs — rank combinations of identical resistors by total resistance, or 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.
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.
[Diagram: phys-circuit] - Available in full study mode
IB-style question — (a) the current in a series circuit
A 6.0 Ω resistor (R₁) and a 3.0 Ω resistor (R₂) are connected in series across a 9.0 V supply (internal resistance negligible). Find the current shown on the ammeter.
Solution
- First find the total resistance with the given series rule:
- Now use the given resistance equation, rearranged for current:
- Put in the numbers (V = 9.0 V, Rs = 9.0 Ω):
Final answer
current I = 1.0 A — and because it is one loop, the ammeter reads this same 1.0 A wherever it sits.
IB-style question — (b) the p.d. across one resistor
For the same series circuit (R₁ = 6.0 Ω, I = 1.0 A), find the p.d. the voltmeter reads across R₁.
Solution
- Use the given resistance equation, rearranged for p.d.:
- Put in the numbers (I = 1.0 A, R₁ = 6.0 Ω):
- Check: R₂ takes the rest — 1.0 × 3.0 = 3.0 V, and 6.0 + 3.0 = 9.0 V ✓
Final answer
voltmeter reads 6.0 V across R₁. The bigger resistor takes the bigger share of the supply.