Compute the equivalent resistance of the network shown in figure and find the current $i$ drawn from the battery.

 The $6\Omega$ and $3\Omega$ resistances are in parallel. Their equivalent resistance is

$\frac{1}{R}=\frac{1}{6}+\frac{1}{3}\text{ or }R=2\Omega$

Now, this $2\Omega$ and $4\Omega$ resistances are in series and their equivalent resistance is $4+2=6\Omega$. Therefore, equivalent resistance of the network $=6\Omega$.

Current drawn from the battery is

$i=\frac{\text{ net emf }}{\text{ net resistance }}=\frac{18}{6}$

$=3A$