Free calculator
Resistors in series and parallel calculator
Resistors in series sit one after another, so the current goes through each in turn and their resistances add. Resistors in parallel sit side by side across the same two points, giving the current more paths, so the total is smaller than any one of them.
Total in parallel, 1 ÷ (1/R1 + 1/R2 + …)
- Nearest E12 value
- 4.7 kΩ
- Same ones in series
- 20 kΩ
The rules
In series, add them up:
In parallel, add the reciprocals:
For exactly two in parallel there's a shortcut, product over sum:
A worked example
You need about 5 kΩ but only have 10 kΩ resistors. Two of them in parallel give 10k × 10k over 10k + 10k = 5 kΩ. Two in series give 20 kΩ.
Free lesson
Now build it yourself
Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.
“Combining resistors” is lesson 6 of 7 in Junction City, Electronics World 2. Free with an account. Lessons unlock in order, so you’ll do the 12 before it first.
Or start from the very first lesson, no account neededThe ideas behind it
- Series vs parallel circuitsOne path or many: what stays the same, what adds up, and how to combine resistors.
- How to calculate equivalent resistanceShrink any series-parallel network to one resistor, one step at a time, and spot what's really in parallel.
- Current divider rule explainedHow a current shares itself between parallel branches, and why the smaller resistor takes the bigger share.
Questions people ask
How do you add resistors in parallel?
Add their reciprocals and take the reciprocal of the sum: 1 ÷ (1/R1 + 1/R2 + …). For just two, R1 × R2 ÷ (R1 + R2) is quicker. Two equal resistors in parallel make half of one.
Why is the parallel total smaller than the smallest resistor?
Each resistor in parallel is another path for current. More paths let more current through at the same voltage, and that's what a lower resistance means.
How do I work out a series-parallel network?
Reduce it a piece at a time: combine any group that's purely in series or purely in parallel into one resistor, redraw, and repeat until one is left.
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