IGCSE Physics · Tool
Circuits
How do you work out resistance in series and in parallel?
In series, resistances add: R = R₁ + R₂ + …, and the same current passes through each resistor. In parallel, 1/R = 1/R₁ + 1/R₂ + …, so the combined resistance is less than the smallest branch, and every branch has the same potential difference. At each resistor, V = I × R.
| Resistor | Resistance | Current | p.d. |
|---|---|---|---|
| R₁ | 2 Ω | 2.0 A | 4.0 V |
| R₂ | 4 Ω | 2.0 A | 8.0 V |
R = R₁ + R₂ = 2 Ω + 4 Ω = 6.0 Ω
I = V ÷ R = 12 V ÷ 6.0 Ω = 2.0 A
V₁ = I × R₁ = 2.0 A × 2 Ω = 4.0 V
V₂ = I × R₂ = 2.0 A × 4 Ω = 8.0 V
The p.d.s add up to the cell’s: 4.0 V + 8.0 V = 12 V.
The current is the same all the way round a series circuit.
Answers are given to 2 significant figures, like the values you typed.
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What to take away
- In series the same current passes through every resistor, and the resistances add.
- In parallel every branch has the same p.d., and the combined resistance is less than the smallest branch.
- At every resistor, the p.d. across it = the current through it × its resistance.
Common mistakes
- Adding resistances in parallel as if they were in series.
- Thinking current is used up as it goes round a series circuit: it is the same everywhere in it.
Questions
How do you work out resistance in series and in parallel?
In series, resistances add: R = R₁ + R₂ + …, and the same current passes through each resistor. In parallel, 1/R = 1/R₁ + 1/R₂ + …, so the combined resistance is less than the smallest branch, and every branch has the same potential difference. At each resistor, V = I × R.
Why is the total resistance in parallel less than the smallest resistor?
Each extra branch gives the current another path, so more current flows for the same p.d. More current for the same p.d. means less resistance overall.
Is the circuit calculator free?
Yes. It runs here in your browser with no account, and every ConceptOrbit course tool is free.
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