Answer
$C_{2}=568~\mu F$
Work Step by Step
Given
$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$
$C=45~\mu F, C_{1}=85~\mu F, C_{2}=?,C_{3}=115~\mu F$
$\frac{1}{45}=\frac{1}{85}+\frac{1}{C_{2}}+\frac{1}{115}$
$\frac{1}{C_{2}}=\frac{1}{45}-(\frac{1}{85}+\frac{1}{115})$
Therefore, $C_{2}=568~\mu F$
How to use the calculator:
$45,*,x^{-1},-,85,*,x^{-1},-,115,*,x^{-1},=,x^{-1},=.$
