In the network shown below, the charge accumulated in the capacitor in steady state will be ........... $\mu C$
JEE MAIN 2023, Medium
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No current will flow in capacitor in steady state, current flowing in the circuit in steady state

$I=\frac{3}{6+4}=\frac{3}{10}$

Potential difference on $6\,\Omega$ resistance

$V=6 \times \frac{3}{10}=1.8 \text { volt }$

Capacitor will have same potential so charge,

$q = CV =(4\,\mu F ) \cdot(1.8 \text { volt })=7.2\,\mu C$

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