As show in the figure, in steady state, the charge stored in the capacitor is....... $\times 10^{-6}\,C$.
JEE MAIN 2022, Diffcult
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$q = CV _{100 \Omega}$
$=\left(1.1 \times 10^{-6}\right)\left(\frac{10}{ R + r } R \right)$
$=1.1 \times 10^{-6}\left(\frac{10}{110} \times 100\right)$
$=10\,\mu C$
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