Question

What total capacitances (in µF) can you make by connecting a 4.45 µF and an 8.10 µF capacitor together?smaller value µFlarger value

Answer 1

We will have the following:

The total capacitances you can make by connecting them in parallel and in series are:

Series:

$\begin{gathered} \frac{1}{C}=\frac{1}{4.45\mu F}+\frac{1}{8.10\mu F}\Rightarrow C=\frac{(4.45\mu F)(8.10\mu F)}{(4.45\mu F)+(8.10\mu F)} \\ \\ \Rightarrow C=2.872111554...\mu F\Rightarrow C\approx2.87\mu F \end{gathered}$

Parallel:

$C=4.45\mu F+8.10\mu F\Rightarrow C=12.55\mu F$

So, the smaller value is:

$2.87\mu F$

And the largest value is:

$12.55\mu F$