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JEE Main 2025 (3 Apr, Shift 2) · Charge Sharing & Loss of Energy

Charge Sharing & Loss of Energy question from JEE Main 2025 (3 Apr, Shift 2)

Q8JEE Main 2025 · 3 APR · SHIFT 2Easy

Charge Sharing & Loss of Energy

Using a battery, a 100 pF capacitor is charged to 60 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V, its capacitance is: (in pF)

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Answer: (d)

Charge $Q = 100 \times 60 = 6000$ pC stays the same.
$6000 = (100 + C_2) \times 20 \Rightarrow 100 + C_2 = 300$
$C_2 = 200$ pF, so the answer is (d).

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