Question

Three capacitors each of capacitance 9 pF are connected in series.

(a) What is the total capacitance of the combination?

(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

Answer 1

(a) Capacitance of each of the three capacitors, C = 9 pF

Equivalent capacitance (C') of the combination of the capacitors is given by the relation, 

`1/"C'" = 1/"C" + 1/"C" + 1/"C"`

`=> 1/"C'" = 1/9 + 1/9 + 1/9 = 1/3`

`=>` C' = 3 pF

Therefore, the total capacitance of the combination is 3 pF.

(b) Supply voltage, V = 120 V

Potential difference (V') across each capacitor is equal to one-third of the supply voltage.

∴ `"V'" = "V"/3 = 120/3` = 40 V

Therefore, the potential difference across each capacitor is 40 V.