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JEE Main 2020 (2 Sep, Shift 2) · Charge Sharing & Loss of Energy

Charge Sharing & Loss of Energy question from JEE Main 2020 (2 Sep, Shift 2)

Q25JEE Main 2020 · 2 SEP · SHIFT 2Easy

Charge Sharing & Loss of Energy

A 10 $\mu$ 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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Answer: (b)

Charge $Q = 10 \times 50 = 500\ \mu$ C stays the same.
$500 = (10 + C_2) \times 20 \Rightarrow 10 + C_2 = 25$
$C_2 = 15\ \mu$ F, so the answer is (b).

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