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2.6) Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination?

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Given, 3 capacitor of 9pF connected in series,

the equivalent capacitance when connected in series is given by 

\frac{1}{C_{equivalent}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}

\frac{1}{C_{equivalent}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3}

C_{equivalent}=3pF

Hence total capacitance of the combination is 3pF.

Posted by

Pankaj Sanodiya

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