JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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)
- A \(600\)
- B \(200\)
- C \(400\)
- D \(100\)
Answer & Solution
Correct Answer
(B) \(200\)
Step-by-step Solution
Detailed explanation
New potential \(=\frac{\mathrm{C}_0 \mathrm{~V}_0}{\mathrm{C}_0+\mathrm{C}}=\frac{\mathrm{V}_0}{3}\)…
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