JEE MAIN - Physics (2025 - 23rd January Evening Shift - No. 24)
At steady state the charge on the capacitor, as shown in the circuit below, is _________ $\mu$C.
_23rd_January_Evening_Shift_en_24_1.png)
Answer
16
Explanation
_23rd_January_Evening_Shift_en_24_2.png)
$\begin{aligned} & \mathrm{i}=\left(\frac{5}{25}\right) \\\\ & \mathrm{Q}=\mathrm{CV} \\\\ & \mathrm{Q}=\left(8 \times 10^{-6}\right)\left(\frac{5}{25} \times 10\right) \\\\ & \mathrm{Q}=\left(\frac{8 \times 5 \times 10^{-2}}{25}\right)=16 \mu \mathrm{C}\end{aligned}$
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