A capacitor of capacity 5 mF is charged to a potential difference of 100V. The battery is disconnected, and then the capacitor is connected in parallel to an uncharged capacitor of capacitance C. the potential difference measured across this combination is 25V. The capacitance C is

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15 mF

5 mF

10 mF

20 mF

answer is A.

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Detailed Solution

Final common potential difference after the two capacitors are connected is
$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} o r, 25 = \frac{5 \times 100 + C (0)}{5 + C} o r, 5 + C = 20 o r, C = 15 m F$

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