There resistors of 3 `Omega` each are connected to a battery of 3 V as shown. Calculate the current drawn from the battery.

Explanation

`R_(4)=R_(1)+R_(2)=3 Omega+3 Omega=6 Omega`
`(1)/(R_(5))=(1)/(R_(4))+(1)/(R_(3))=(1)/(6) +(1)/(3)=(1+2)/(6)=(3)/(6) =(1)/(2) Omega`
`R_(5) = 2Omega`
Current `I = (V)/(R_(5))=(3 V)/(2 Omega)= 1.5` Amp