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A 10 $μ$ F capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 volts. The capacitance of second capacitor is

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15 $μ$ F

30 $μ$ F

20 $μ$ F

10 $μ$ F

answer is A.

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Detailed Solution

Here, $q_{1} = 10 \times 50 = 500 μC$
$C_{1} = 10 μF, C_{2} = ? and q_{2} = 0$
Now $V = \frac{q_{1} + q_{2}}{C_{1} + C_{2}} = \frac{500 + 0}{C_{1} + C_{2}}$
or $C_{1} + C_{2} = \frac{500}{20} = 25 μF$
$∴ C_{2} = 25 - C_{1} = 25 - 10 = 15 μF$

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