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Series and parallel circuits
How resistors combine and where current and voltage divide.
How resistors combine and where current and voltage divide.
Parts joined end to end, so one path runs through all of them, are in series. Parts joined across the same two points, so the current has a choice of paths, are in parallel. Every larger circuit is built from these two patterns.
| Series | Parallel | |
|---|---|---|
| Current | the same in every part | splits; the branches add up to the total |
| Voltage | splits; the drops add up to the supply | the same across every branch |
| Total resistance | R₁ + R₂ + … (always larger) | smaller than the smallest branch |
The reasons follow from conservation, covered in Kirchhoff's laws: in series there is nowhere for current to go but through the next part, and in parallel each branch is connected straight to the supply.
Series resistances simply add: 100 Ω + 220 Ω = 320 Ω. Parallel ones are easier to understand as extra paths. Each added branch gives current another way through, so the total resistance falls.
For two branches, R = (R₁ × R₂) ÷ (R₁ + R₂): 47 Ω and 100 Ω in parallel give 4700 ÷ 147 = 32 Ω. Two equal resistors give half of one: 100 Ω ∥ 100 Ω = 50 Ω. For more branches, add the reciprocals: 1/R = 1/R₁ + 1/R₂ + …; 10, 20 and 50 Ω give 1/10 + 1/20 + 1/50 = 0.17, so 5.88 Ω. A quick check: the answer must be smaller than your smallest branch.
Mixed circuits are solved by reducing: replace each parallel group by its equivalent, then add what is in series. A 100 Ω in series with two 100 Ω in parallel is 100 + 50 = 150 Ω.
Two resistors in series share the supply in proportion to their size. The current is E ÷ (R₁ + R₂), and each resistor drops its own R × I, so V₂ = E × R₂ ÷ (R₁ + R₂). With 12 V across 30 Ω and 10 Ω the current is 0.3 A and the drops are 9 V and 3 V.
A divider gives a lower voltage, but only if the load takes little current. Connect a load across R₂ and it is in parallel with it, lowering R₂ and the output. Two 10 kΩ resistors from 12 V give 6 V; a 10 kΩ load on the lower one makes it 5 kΩ and the output falls to 4 V. The same effect makes a low-resistance voltmeter read low.
In parallel, current favours the easier path: branch currents are in inverse proportion to resistance. A total of 3 A into 10 Ω and 20 Ω parallel branches splits 2 A and 1 A, because the 10 Ω branch is half the resistance and takes twice the current.