Three capacitors each of capacitance 9pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potantial difference across each capacitor if the combination is connected to a 120 V supply?

Explanation

Given `C_(1)=C_(2)=C_(3)=9xx10^(-12)pF`
(a) For a series of combination of 'n' identical capacitors, their equivalent capacitance
`C_(S)=(C)/(n)=(9xx10^(-12))/(3)=3xx10^(-12)F`
i.e., `C_(S)=3pF`
(b) `V_(1):V_(2):V_(3)::(1)/(C_(1)):(1)/(C_(2)):(1)/(C_(3))`
i.e., `V_(1):V_2:V_(3)::(1)/(C):(1)/(C):(1)/(C)`
i.e., `V_(1):V_(2):V_(3)::1:1:1`
`therefore V_(1)=V_(2)=V_(3)=(1)/(3)xx120V`
`V_(1)=V_(2)=V_(3)=40V`.
Thus voltage across each of the capacitance is 40 V.