Answer
$3.71\times10^{-6}F$
Work Step by Step
When capacitors are wired in series, we use equation 19–6 to find the equivalent capacitance.
$$\frac{1}{C_{eq}}=\frac{1}{3.00\times10^{-6}F }+\frac{1}{4.00\times10^{-6}F} $$
$$C_{eq}=1.714\times10^{-6}F $$
This combination is then wired in parallel with a third capacitor, so we use equation 19–5 to find the final equivalent capacitance.
$$C_{eq}=1.714\times10^{-6}F +2.00\times10^{-6}F $$
$$=3.71\times10^{-6}F$$
