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?

Updated On: Aug 26, 2024
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Solution and Explanation

(a) Capacitance of each of the three capacitors, C = 9 pF Equivalent capacitance (Ceq) of the combination of the capacitors is given by the relation, \(\frac{1}{Ceq}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=\frac{3}{C}=\frac{3}{9}=\frac{1}{3 }=\frac{1}{C_eq}=\frac{1}{3} c_eq=3pF\)  

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.

\(V1=\frac{V}{3} =\frac{120}{3} =40 V\)

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

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Concepts Used:

Combination of Capacitors

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