Question

Find the total capacitance for three capacitors connected in series, given their

individualcapacitances are 2.000nF, 5.000pF, and 8.000 μF.

Answer 1

$$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$

$$\frac{1}{C}=\frac{1}{2.000*10^{-9}}+\frac{1}{5.000*10^{-12}}+\frac{1}{8.000*10^{-6}}$$

$$C=4.988*10^{-12}\:\rm F=4.988\: pF$$