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Three capacitors each of capacitance $9 \;pF$ are connected in series. What is the total capacitance of the combination?

$\begin{array}{1 1} 3\;\mu F \\ 6\;\mu F \\ 5\;\mu F \\ 2\;\mu F \end{array} $

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A)
Solution :
Capacitance of each of the three capacitors, $C = 9\; pF $
Equivalent capacitance $(C_{eq})$ of the combination of the capacitors is given by the relation,
$\large\frac{1}{C_{eq}} =\frac{1}{C}+\frac{1}{C} +\frac{1}{C} =\frac{3}{C} =\frac{3}{9}=\frac{1}{3}$
=> $\large\frac{1}{C_{eq}} =\frac{1}{3} => $$C_{eq}=3 \mu F$
Therefore, total capacitance of the combination is $3 \mu F$.
 

 

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