Question

Calculate the total capacitance of the three capacitors 30µF, 20µF & 12µF connected in series across a d.c. supply

Answer 1

Consider that three capacitors connected in series have the following total capacitance:

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

where,

C1 = 30µF

C2 = 20µF

C3 = 12µF

Consider that the LCM of the three previous numbers is 60 (to sum the fractions).

Replace the previous values of the parameters into the formula for C and simplify:

$\begin{gathered} \frac{1}{C}=\frac{1}{30}+\frac{1}{20}+\frac{1}{12} \\ \frac{1}{C}=\frac{2+3+5}{60}=\frac{10}{60}=\frac{1}{6} \\ C=6 \end{gathered}$

Hence, the total capacitance is 6µF