# please give solution fastq44

Under steady state condition, there will be no current through the capacitor arm.

Equivalent resistance between A and B:

${R}_{AB}=\frac{3\times 2}{3+2}=\frac{6}{5}=1.2\Omega $

So, the current in the circuit.

$I=\frac{6}{1.2+2.8}=\frac{6}{4}=1.5A$

Potential drop across AB:

${V}_{AB}=I\times {R}_{AB}=1.5\times 1.2=1.8V$

So the current through the 2ohm resistor

$i=\frac{1.8}{2}=0.9A$

Regards

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