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Three capacitors of capacity $C_1, C_2 C_3$ are connected in series. Their total capacity will be
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Verified Answer
The correct answer is:
$\left(C_1^{-1}+C_2^{-1}+C_3^{-1}\right)^{-1}$
$\frac{1}{C_R}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \Rightarrow C_R=\left(C_1^{-1}+C_2^{-1}+C_3^{-1}\right)^{-1}$
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