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Question

A 5μF capacitor is fully charged by a 12 V battery and then disconnected. If it is connected now parallel to an uncharged capacitor, the voltage across it is 3 V. Then, the capacity of the uncharged capacitor is

A
5μF
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B
25μF
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C
50μF
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D
10μF
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E
15μF
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Solution

The correct option is B 25μF
Given, C=5μF
and V=12V
We know that, Q=CV
Q=5×12=60μC
According to the question,
60C=3
C=20μF
Net capacitance =20+5=25μF.

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