A 5μF capacitor is fully charged by a 12V battery and then disconnected. If it is connected now parallel to an uncharged capacitor, the voltage across it is 3V. 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 B25μF 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.