List the resistor values separated by spaces, commas or semicolons, choose series or parallel, and read the equivalent resistance. Prefixes work, so 4.7k means 4,700 ohms.
Series: R = R1 + R2 + R3 + …
Parallel: 1/R = 1/R1 + 1/R2 + 1/R3 + …
Three resistors of 1 kΩ, 2.2 kΩ and 470 Ω in parallel: 1/1000 + 1/2200 + 1/470 = 0.0035822 S, so the total is about 279.2 Ω.
Add the values. Three resistors of 1 kΩ, 2.2 kΩ and 470 Ω in series give 1000 + 2200 + 470 = 3670 Ω, shown as 3.67 kΩ.
Add the reciprocals and take the reciprocal of the sum: 1/R = 1/R1 + 1/R2 + … For two equal 100 Ω resistors this is 1/100 + 1/100 = 1/50, so 50 Ω. The result is always lower than the smallest resistor.
Every extra path lets more current through at the same voltage, so the combined resistance falls.
Yes. Try combinations, such as two resistors in parallel or series, until the total is close to the value you need. Tolerance of the real parts changes the result.
It is 1 ÷ the total resistance, in amps. It is a quick way to see how much current the network would draw from 1 V; multiply by your actual voltage for the real current.