Question

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

1. What is the total capacitance of the combination?

2. What is the potential difference across each capacitor, if the combination is connected to a 120 volt supply?

Answer 1

Given C₁ = C₂ = C₃ = 9pF = 9 × 10^-12 F; V= 120 volt.

1. Total capacitance of the series combination is given by

\(\frac{1}{c}\) = \(\frac{1}{c_1}+\frac{1}{c_2}+\frac{1}{c_3}\)

\(\frac{1}{c}\) = 3x\(\frac{1}{9\times10^{-12}}\) = \(\frac{1}{3\times10^{-12}}\)

∴ C = 3 × 10^-12 F = 3pF.

2. Let q be the charge on each capacitor. Then, sum of the potential difference across their plates must be equal to 120 V.

ie. V₁ + V₂ + V₁ = 120

or \(\frac{q}{c_1}\) + \(\frac{q}{C_2}\) + \(\frac{q}{c_3}\) = 120

or \(\frac{3q}{9\times10^{-12}}\) = 120

or q = 360 × 10^-12 C

Since, all the capacitors are of same capacitance,

= \(\frac{q}{capacitance}\) = \(\frac{360\times10^{-12}}{9\times10^{-12}}\) = 40 V