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

Explanation

Two combinations of two parallel resistors of `10 Omega` each connected in series.
`(1)/(R_(5))=(1)/(R_(1))+(1)/(R_(2))=(1)/(10)+(1)/(10)= (1+1)/(10) = (2)/(10)`
`(2)/(10)=(1)/(5) Omega`
`R_(5) = 5 Omega`
`(1)/(R_(6))=(1)/(R_(3))+(1)/(R_(4))=(1)/(10)+(1)/(10)=(1+1)/(10)=(2)/(10)`
`R_(6) =5 Omega`
`R_(eq)=R_(5) +R_(6)= 5 Omega+5 Omega = 10 Omega`
`I = (V)/(R) = (3)/(10)= 0.3 ` Amp.