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

Explanation

The `6Omega and 3 Omega` resistance are in parallel. Their equivalent resistance is `1/R=1/6+1/3 or R=2Omega`

Now, this `2Omega and 4Omega` resistanes are in series and their equivalent resistance is `4+2=6Omega`. Therefore, equivalent reistance of the network `=6 Omega`.

Current drawn from the battery is
`i=("netemf")/("net resistance")=18/6`
`=3A`