A 47kΩ resistor and 39 kΩ resistor are in parallel

Question

A 47kΩ resistor and 39 kΩ resistor are in parallel, and the pair in series with a 22 kΩ resistor. What is

the resistance of the combination?

Answer 1

Let R1 = 47 kΩ, R2 = 39 kΩ, and R3 = 22 kΩ. The circuit would look like this:

1/R_{eq 1-2} = 1/R₁ + 1/R₂ = 1/47000 + 1/39000

R_{eq 1-2} = 1/(1/47000 + 1/39000) = 21314 Ω

R_{eq 1-2-3} = R_{eq 1-2} + R₃ = 21314 + 22000 = 43314 Ω which need to be rounded to an appropriate number of sig figs