A $10 μ F$ capacitor is fully charged to a potential difference of 50V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20V. The capacitance of the second capacitor is:

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$10 μ F$

$15 μ F$

$20 μ F$

$30 μ F$

answer is A.

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Detailed Solution

$V_{C M} (c o m m o n p o t e n t i a l) = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$

$\begin{array}{l} \Rightarrow 20 = \frac{10 \times 50}{10 + C_{2}} (∵ F o r u n c h a r g e d c a p a c i t o r, V = 0) \\ \Rightarrow 10 + C_{2} = 25 \\ \Rightarrow C_{2} = 15 μ F \end{array}$

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