Chapter 6 - Section 6.9 - Reciprocal Formulas Using a Calculator - Exercises - Page 261: 16

$R_{3}=4590~\Omega$.

Given $\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$ $R=1830~\Omega, R_{1}=4560~\Omega, R_{2}=9150~\Omega,R_{3}=?$ $\frac{1}{1830}=\frac{1}{4560}+\frac{1}{9150}+\frac{1}{R_{3}}$ $\frac{1}{R_{3}}=\frac{1}{1830}-(\frac{1}{4560}+\frac{1}{9150})$ Therefore, $R_{3}=4590~\Omega$ How to use the calculator: $1830*x^{-1}-4560*x^{-1}-9150*x^{-1}=x^{-1}=$