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Q.

A $10 μ F$ capacitor is fully charged to a potential difference of 50 V. After removing the source voltage, it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is:

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a

$15 μ F$

b

$30 μ F$

c

$20 μ F$

d

$10 μ F$

answer is A.

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Detailed Solution

Given,
Capacitance of capacitor, $C_{1} = 10 μ F$
Potential difference before removing the source voltage, $V_{1} = 50 V$
If $C_{2}$ be the capacitance of uncharged capacitor, then common potential is
$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} \Rightarrow 20 = \frac{10 \times 50 + 0}{10 + C} \Rightarrow C = 15 μ F$

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A 10μF capacitor is fully charged to a potential difference of 50 V. After removing the source voltage, it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is: