What is the steady state current in the $2\Omega$ resistor shown in fig. The internal resistance of the battery is negligible and the capacitance $C=0.5\mu F$

In steady state, current through capacitor is zero. 

Therefore, current in the circuit becomes 

$i=\frac{V}{r+R_{eff}}=\frac{2}{1.2+2.8}=1.5A$

Current through $2\Omega$ resistor is 

${\mathrm{i}}_1=\frac{3}{3+2}\times 1.5\mathrm{A}=0.9\mathrm{A}$