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A 10 μF capacitor is charged to a potential difference 50 V and it is then connected to another uncharged capacitor in parallel. If common potential difference becomes 20 V, then capacitance of the second capacitor is
  • a)
    10 μF
  • b)
    15 μF
  • c)
    20 μF
  • d)
    30 μF
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A 10 μF capacitor is charged to a potential difference 50 V and it is ...
Q=10×50=500μC

Second case:

500=(C+10)×20

⇒C=15μF
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