Question 19.9: Find the total capacitance for three capacitors connected in...

Entering the given capacitances into the expression for \frac{1}{C_{S}} gives \frac{1}{C_{S}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}

\frac{1}{C_{ S }}=\frac{1}{1.000 \mu F }+\frac{1}{5.000 \mu F }+\frac{1}{8.000 \mu F }=\frac{1.325}{\mu F }                   (19.64)

Inverting to find C_{S} yields C_{ S }=\frac{\mu F }{1.325}=0.755 \mu F.

Discussion
The total series capacitance C_{S} is less than the smallest individual capacitance, as promised. In series connections of capacitors, the sum is less than the parts. In fact, it is less than any individual. Note that it is sometimes possible, and more convenient, to solve an equation like the above by finding the least common denominator, which in this case (showing only whole-number calculations) is 40. Thus,

\frac{1}{C_{S}}=\frac{40}{40 \mu F }+\frac{8}{40 \mu F }+\frac{5}{40 \mu F }=\frac{53}{40 \mu F },              (19.65)

so that

C_{ s }=\frac{40 \mu F }{53}=0.755 \mu F.                   (19.66)