Calculate the current the current drawn from the battery by the network of resistors shown in figure.

Explanation

The given network is drawn as
Here `R_(1)/R_(4)=1/2 and R_(5)/R_(3)=2/4=1/2`
`:. R_(1)/R_(4)=R_(5)/R_(3)`
According to wheatstone bridge, no current will be flow in R_(2) resistor.
`:. R_(1)` and `R_(5)` are in series,
`:. R_(15)=1+2=3 Omega` and `R_(4)` and `R_(3)` are in series
`:. R_(43)=2+4+6 Omega`
Now `R_(15)` and `R_(43)` are parallel, the equivalent resistance R is given by
`1/R=1/3+1/6=(2+1)/6=3/6=1/2`
`:. R=2Omega`.
`:.` Current drawn from the battery `I=V/R=4/2=2 A." "[ :' V=IR]`