Physics - Current Electricity Question with Solution | TestHub

The arrangement is shown in figure.
Here, two reisistance of $1\Omega$ and $2\Omega$ are in series, which form $3\Omega$ which is in parallel with $3\Omega$ resistance. Therefore, the effective resistance $\frac{(1+2)\times 3}{(1+2)+3}=\frac{3}{2}\Omega$
$\therefore$ Current in the circuit, $$\begin{array}{l}\mathrm{I}=\frac{3}{(3/2)}=2\mathrm{A}\\ \therefore \text{ Current in }3\Omega \text{ resistor }=\frac{\mathrm{I}}{2}=\mathrm{IA}\end{array}$$