JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2025
- Medium
- MCQ
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)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(200\)
Step-by-step Solution
Detailed explanation

New potential \(=\frac{\mathrm{C}_0 \mathrm{~V}_0}{\mathrm{C}_0+\mathrm{C}}=\frac{\mathrm{V}_0}{3}\)
\(\begin{aligned} & 3 \mathrm{C}_0 \mathrm{~V}_0=\mathrm{C}_0 \mathrm{~V}_0+\mathrm{CV}_0 \\ & 2 \mathrm{C}_0 \mathrm{~V}_0=\mathrm{CV}_0 \\ & \mathrm{C} \Rightarrow 2 \mathrm{C}_0\end{aligned}\)
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