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:

`(1)/(C)=(1)/(C_1)+(1)/(C_2)+(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} (1)/(C)=(1)/(30)+(1)/(20)+(1)/(12) \\ (1)/(C)=(2+3+5)/(60)=(10)/(60)=(1)/(6) \\ C=6 \end{gathered}`

Hence, the total capacitance is 6µF