Answer
a. $28.8\times10^{-6}F$
b. $8.0\times10^{-7}F $
Work Step by Step
a. When capacitors are wired in parallel, we use equation 19–5 to find the equivalent capacitance.
$$C_{eq}=4.8\times10^{-6}F +4.8\times10^{-6}F+ 4.8\times10^{-6}F+4.8\times10^{-6}F+4.8\times10^{-6}F+4.8\times10^{-6}F $$
$$=28.8\times10^{-6}F$$
b. When capacitors are wired in series, we use equation 19–6 to find the equivalent capacitance.
$$\frac{1}{C_{eq}}=\frac{1}{4.8\times10^{-6}F }+\frac{1}{4.8\times10^{-6}F }+\frac{1}{4.8\times10^{-6}F} +\frac{1}{4.8\times10^{-6}F} +\frac{1}{4.8\times10^{-6}F} +\frac{1}{4.8\times10^{-6}F}$$
$$C_{eq}=8.0\times10^{-7}F $$
