JEE MAIN - Physics (2025 - 3rd April Evening Shift - No. 8)
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 )
600
100
400
200
Explanation
_3rd_April_Evening_Shift_en_8_1.png)
$$\begin{aligned} & \text { New potential }=\frac{\mathrm{C}_0 \mathrm{~V}_0}{\mathrm{C}_0+\mathrm{C}}=\frac{\mathrm{V}_0}{3} \\ & 3 \mathrm{C}_0 \mathrm{~V}_0=\mathrm{C}_0 \mathrm{~V}_0+\mathrm{CV}_0 \\ & 2 \mathrm{C}_0 \mathrm{~V}_0=\mathrm{CV}_0 \\ & \mathrm{C} \Rightarrow 2 \mathrm{C}_0 \end{aligned}$$
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