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

Explanation

Resultant resistance of `R_(1), R_(2) and R_(3)` :
`R'=R_(1) + R_(2) + R_(3)`
`=10+10+10`
`=30 Omega`
Resultant resistance of the circuit=R:
`(1)/(R)=(1)/(R')+(1)/(R_(4)) =(1)/(30) + (1)/(10)=(1+3)/(30)=(4)/(30)`
`R=(30)/(4)=(15)/(2)=7.5 Omega`
`V=3` volts`" "` I=?
V=IR
`implies I=(V)/(R)=(3)/(7.5) =(30)/(75)=0.4 A`