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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 second capacitor is

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$30 μ F$

$20 μ F$

$15 μ F$

$10 μ F$

answer is C.

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Detailed Solution

$C_{1} = 10 μ F$ (Given)Potential difference before removing the source voltage , V₁ = 50 V.

If C₂ be the capacitance of uncharged capacitor, then common potential is

$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} \Rightarrow 20 = \frac{10 \times 50 + 0}{10 + C} ∴ C = 15 μF$

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