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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)

  1. Charge Q=100×60=6000Q = 100 \times 60 = 6000 pC stays the same.
  2. 6000=(100+C2)×20⇒100+C2=3006000 = (100 + C_2) \times 20 \Rightarrow 100 + C_2 = 300
  3. C2=200C_2 = 200 pF, so the answer is (d).
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