Ncert Exercises & Solutions

Three capacitors, each having a capacitance of $9~\text{pF},$ are connected in series and then connected to a $120~\text V$ supply. What is the total capacitance of the combination and the potential difference across each capacitor, respectively?
1. $27~\text{pF}~\text{and}~30~\text V$
2. $3~\text{pF}~\text{and}~120~\text V$
3. $3~\text{pF}~\text{and}~40~\text V$
4. $27~\text{pF}~\text{and}~120~\text V$

(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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