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_eq) of the combination of the capacitors is given by the relation, 1/C_eq=1/C+1/C+1/C=3/C=3/9=1/3
⇒1/C_eq=1/3⇒C_eq=3 pF

Therefore, total capacitance of the combination is 3 pF.
(b) Supply voltage, V = 100 V
Potential difference (V₁) across each capacitor is equal to one-third of the supply voltage. ∴V₁=V/3=120/3=40V
Therefore, the potential difference across each capacitor is 40 V.