Answer
$C_{2}=1.30*10^{-8}F$.
Work Step by Step
Given
$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$
$C=4.45*10^{-9}F, C_{1}=5.08*10^{-8}F, C_{2}=?, C_{3}=7.79*10^{-9}F$
$\frac{1}{4.45*10^{-9}}=\frac{1}{5.08*10^{-8}}+\frac{1}{C_{2}}+\frac{1}{7.79*10^{-9}}$
$\frac{1}{C_{2}}=\frac{10^{9}}{4.45}-(\frac{10^{8}}{5.08}+\frac{10^{9}}{7.79})$
$\frac{1}{C_{2}}=10^{8}(\frac{10}{4.45}-(\frac{1}{5.08}+\frac{10}{7.79}))$
$\frac{1}{C_{2}}=10^{8}(\frac{1}{0.445}-(\frac{1}{5.08}+\frac{1}{0.779}))$
Therefore, $C_{2}=1.30*10^{-8}F$
How to use the calculator:
$0.445*x^{-1}-5.08*x^{-1}-0.779*x^{-1}=x^{-1}=$
