Answer to Question #312054 in Physics for Bibe

Question

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

individual capacitances are 1.000, 5.000, and 8.000 µF. (0.755µF)

Answer 1

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

$$\frac{1}{C}=\frac{1}{1.000\:\rm \mu F}+\frac{1}{5.000\:\rm \mu F}+\frac{1}{8.000\:\rm \mu F}=1.325$$

$$C=0.755\:\rm \mu F$$