Ncert Exercises & Solutions

Q 2.6) Three capacitors connected in series have capacitance of 9pF each.

(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 V supply?

(1) Given,

The capacitance of the three capacitors, C = 9 pF

Equivalent capacitance (c_eq) is the capacitance of the combination of the capacitors given by

$\begin{array}{l} \frac{1}{C_{e q}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} = \frac{3}{9} = \frac{1}{3} \\ \Rightarrow \frac{1}{C_{e q}} = \frac{1}{3} \Rightarrow C_{e q} = 3 p F \end{array}$

Therefore, the total capacitance = 3pF.

(2) Given, supply voltage, V = 100v

The potential difference (V₁) across the capacitors will be equal to one–third of the supply voltage. $∴ V_{1} = \frac{V}{3} = \frac{120}{3} = 40 V$

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

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