Answer
$R_{2}=321\Omega$
Work Step by Step
Given
$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$
$R=90\Omega, R_{2}=?, R_{1}=125\Omega$
$\frac{1}{90}=\frac{1}{125}+\frac{1}{R_{2}}$
$\frac{1}{R_{2}}=\frac{1}{90}-\frac{1}{125}$
Therefore, $R_{2}=321\Omega$
How to use the calculator:
$90,*,x^{-1},-,125,*,x^{-1},=,x^{-1},=.$
