Answer
$R_{2}=3240~\Omega$.
Work Step by Step
Given
$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$
$R=1250~\Omega, R_{1}=3750~\Omega, R_{2}=?,R_{3}=4450~\Omega$
$\frac{1}{1250}=\frac{1}{3750}+\frac{1}{R_{2}}+\frac{1}{4450}$
$\frac{1}{R_{2}}=\frac{1}{1250}-(\frac{1}{3750}+\frac{1}{4450})$
Therefore, $R_{2}=3240~\Omega$
How to use the calculator:
$1250*x^{-1}-3750*x^{-1}+4450*x^{-1}=x^{-1}=$
