How it’s calculated
In series the same current flows through every resistor, so resistances add. In parallel every resistor sees the same voltage, so conductances (1/R) add and the total is always smaller than the smallest resistor.
Example: 100 Ω and 220 Ω in parallel give 1 ÷ (1/100 + 1/220) = 22,000 ÷ 320 = 68.75 Ω. In series they give 320 Ω.
Frequently asked questions
What is the shortcut for two resistors in parallel?
Product over sum: R = R1 × R2 ÷ (R1 + R2). For 100 Ω and 220 Ω that is 22,000 ÷ 320 = 68.75 Ω.
What about identical resistors in parallel?
n identical resistors R in parallel give R ÷ n. Two 1 kΩ resistors in parallel make 500 Ω.
How do I handle a mixed series-parallel network?
Reduce it step by step: combine each parallel group to one equivalent, then add the series parts, and repeat.
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Sources
Formulas are taken from the free public references above. Results are provided “as is” for informational and educational purposes only. See our disclaimer.
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