Calculate the current drawn from the battery in the given network .

Explanation

The circuit can be redrawn as given in Fig. It is a Wheatstone bridge arrangement in which `R_1/R_5 = R_4/R_3` i.e., the bridge is balanced one. Hence resistance `R_2 = 5Omega` is superfluous and may be ignored.
` therefore` Value of resistance` R_15` of series combination of `R_1` and `R_5 = 1 + 2 =3Omega`

Value of resistance `R_43` of series combination of `R_4` and `R_3 = 2 +4=6Omega`
` therefore ` Net resistance of network R = resistance of parallel grouping of `R_15` and `R_43`
`R = (R_15 xx R_43)/(R_15 xx R_43) = (3 xx 6)/(3+6) = 2 Omega`
` therefore ` Current drawn from the battery in the given network `I = epsi/R = (4V)/(2Omega) = 2A`