Answer
Maximum $=1.90\times10^{-8}F$
Minimum $=1.7\times10^{-9}F $
Work Step by Step
The maximum capacitance is achieved when the 3 capacitors are wired in parallel. Use equation 19–5 to find the equivalent capacitance.
$$C_{eq}=3200\times10^{-12}F +5800\times10^{-12}F+ 0.0100\times10^{-6}F $$
$$=1.90\times10^{-8}F$$
The minimum capacitance is achieved when the 3 capacitors are wired in series. Use equation 19–6 to find the equivalent capacitance.
$$\frac{1}{C_{eq}}=\frac{1}{3200\times10^{-12}F }+\frac{1}{5800\times10^{-12}F }+\frac{1}{0.0100\times10^{-6}F} $$
$$C_{eq}=1.7\times10^{-9}F $$
