A 10μF 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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15μF

10μF

30μF

20μF

answer is A.

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Detailed Solution

VCMcommon potential=C1V1+C2V2C1+C2⇒20=10×5010+C2∵For uncharged capacitor, V=0⇒10+C2=25⇒C2=15μF

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