How to Calculate Total Resistance of Resistors in Parallel

For resistors in parallel the total is the reciprocal of the sum of reciprocals, 1 ÷ (1/R₁ + 1/R₂ + …), which is always smaller than the smallest resistor. List the resistances separated by commas.

Explanation

Resistors joined side by side, in parallel, give the charge more than one route. To combine them, add the reciprocals and then take the reciprocal of that sum.

\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots + \dfrac{1}{R_n}

$R$ is the total resistance, in ohms (Ω)
$R_1, R_2, \ldots$ are the individual resistances, in ohms (Ω)

In parallel every resistor sees the same voltage, and it is the current that splits: $\frac{V}{R_1}$ down one branch, $\frac{V}{R_2}$ down the next. Adding routes makes it easier for current to flow overall.

Example

The defaults are three resistors of 10, 20 and 30 Ω.

\dfrac{1}{R} = \dfrac{1}{10} + \dfrac{1}{20} + \dfrac{1}{30} = \dfrac{11}{60}

R = \dfrac{60}{11} \approx 5.4545\,\Omega

The total resistance is about 5.4545 Ω, from a count of 3.

Notes

A parallel total is always smaller than the smallest resistor in the group, here smaller than 10 Ω. That is the sharpest contrast with a series connection, and it doubles as a sanity check: an answer above 10 Ω would mean a slip somewhere.

Do not stop at the sum of the reciprocals. You still have to invert it at the end.

For $n$ identical resistors of $R_0$ , the total is $\frac{R_0}{n}$ . Two 10 Ω resistors in parallel make 5 Ω.

Zero and negative resistances are rejected.

The sockets in a house are wired in parallel, which is why plugging in more appliances draws more current from the supply.