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

Explanation

The equivalent circuit for the given network is shown in figure, which is a Wheatstone bridge. Here `P = 1 Omega, Q=2Omega, R= 2 Omega` and `S=4 Omega`

and `(1 Omega)/(2 Omega) = (2 Omega)/( 4 Omega)` i.e., `P/Q = R/S`
Thus the bridge is a balanced Wheatstone bridge. The resistance `5 Omega` in arm BD is ineffective. Therefore, the equivalent circuit reduces to the circuit.

The resistance of arm ABC `= 1 + 2 = 3 Omega`
Resistance of arm ADC `= 2 +4 = 6 Omega`
The resistance of thses two arms are in parallel. The equivalent resistance `R_(eq)` between A and C is
`R_(eq) = ( 3 xx 6)/(3 +6) = 2 Omega`
Current `I =V/R_(eq) = 4/2 =2 A`