Explanation

(b) Required arrangement is shown in figure.
The equivalent circuit will look like (since the two resistances of 1$\Omega$ and 2$\Omega$ are in series, which from 3$\Omega$ which is in parallel with 3$\Omega$ resistance).
Therefore, the effective resistance is
$\frac{(1 + 2) \times 3}{(1 + 2) + 3} = \frac{3}{2}\Omega$
∴ Current in the circuit,
$I = \frac{3}{\left( \frac{3}{2} \right)} = 2A$
∴ Current in 3$\Omega$ resistor= $\frac{I}{2} = 1A$