Add the inverses and invert the result: R = 1 ÷ (1 ÷ R1 + 1 ÷ R2 + ...). For two, R = R1 × R2 ÷ (R1 + R2). Example: 220 Ω and 330 Ω give 132 Ω. The result is always smaller than the smallest resistor.
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Add resistors in series and parallel, solve mixed circuits and find which standard values to combine to reach the resistor you need
In series resistors form a single path and the resistance adds: 100 Ω, 220 Ω and 330 Ω give 650 Ω. In parallel the current splits and what adds is the inverse: 1 ÷ R = 1 ÷ R1 + 1 ÷ R2 + ..., and the result is always smaller than the smallest resistor. For two resistors, R = R1 × R2 ÷ (R1 + R2): 220 Ω in parallel with 330 Ω gives 132 Ω. For N equal resistors it is the value divided by N: two 10 kΩ in parallel give 5 kΩ. The tool computes both combinations for a list of up to 20 values, shows the voltage, current and power of each resistor when you enter a voltage, and evaluates mixed circuits such as (1k | 2.2k) + 470. The third tab works backwards: given the value you need and the series you have at home (E6, E12, E24, E48 or E96), it tries every combination of two or three resistors in series, in parallel and mixed, and lists the ones that come closest.
Series: R = R1 + R2 + ... + Rn. With 100, 220, 330 and 470 Ω the sum is 1,120 Ω; on 12 V, the same current flows through all of them, 10.71 mA, and the voltage divides in proportion to value: 1.07 V, 2.36 V, 3.54 V and 5.04 V, which add up to 12 V. Parallel: R = 1 ÷ (1 ÷ R1 + 1 ÷ R2 + ... + 1 ÷ Rn). With 1 kΩ, 2.2 kΩ and 4.7 kΩ the result is 599.8 Ω, below the smallest of them. In parallel all receive the same voltage, and the smallest value takes the largest current.
A mixed circuit is solved from the inside out. In (1 kΩ ∥ 2.2 kΩ) + 470 Ω, the parallel part gives 1,000 × 2,200 ÷ 3,200 = 687.5 Ω, and adding the 470 Ω in series gives 1,157.5 Ω. To write this in the tool, use | for parallel and + for series: parallel is computed before the sum, like multiplication before addition, and parentheses change the order.
Standard series such as E12 and E6 have few values per decade: E12 goes from 2.7 kΩ straight to 3.3 kΩ, with no 3 kΩ. To reach 3 kΩ with E12 the tool finds 1.2 kΩ + 1.8 kΩ, 1.5 kΩ + 1.5 kΩ (two equal ones in series) and 3.3 kΩ ∥ 33 kΩ, all exact. In E6, which has neither 1.2 kΩ nor 1.8 kΩ, 1.5 kΩ + 1.5 kΩ remains. 50 Ω comes from two 100 Ω in parallel, and 75 Ω from two 150 Ω.
Nominal values are only the start: 5% and 10% resistors add their own deviation. An arrangement of two 5% resistors can be off by about 5% in total, on top of the rounding error shown. When accuracy matters, look for a 1% part of the value you want instead of combining 5% parts.
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Add the inverses and invert the result: R = 1 ÷ (1 ÷ R1 + 1 ÷ R2 + ...). For two, R = R1 × R2 ÷ (R1 + R2). Example: 220 Ω and 330 Ω give 132 Ω. The result is always smaller than the smallest resistor.
The math runs in your browser; no value is sent to a server. Combinations use nominal values: resistor tolerance adds on top of the error shown.