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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