Physics - Current Electricity Question with Solution | TestHub

Resistance in uppermost branch $R\text{ = }\frac{2\times 3}{2+3}=\frac{6}{5}=1.2 \mathrm{\Omega }$

After a long time, branch with the capacitor can be considered open circuited

Hence,

$R_{eq}\text{ = 2.8+1.2 = 4}$

$I\text{= }\frac{V}{R}=\frac{6}{4}=\frac{3}{2} \mathrm{A}$

$I_{2\mathrm{\Omega }}\text{= }\frac{I{R}_2}{R_1+R_2}=\frac{\frac{3}{2}\times 3}{5} \mathrm{A}$

$I_{2\mathrm{\Omega }}\text{ = }\frac{9}{10}=0.9\text{ A}$