A uniform wire of resistance `12 Omega` is cut into three pieces so that the ratio of the resistances `R_(1) : R_(2) : R_(3) = 1 : 2 : 3` and the three pieces are connected to form a triangle across which a cell of emf 8 V and internal resistance `1 Omega` is connected as shown.
Calculate the current through each part of the circuit.

Explanation

`R_(1) + R_(2) + R_(3) = 12 Omega`
`(1 +2 +3)x = 12 Omega`
`rArr x= 2 Omega`
Net resistance of the combination of resistances,
`R_(AC) = (R.R_(3))/(R_(3) + R.)`
`R_(AC) = (36)/(12) = 3Omega`
Current in this network
`I= (E )/(R_(AC) +r) = (8)/(3+1) = 2A`
Voltage across the branch `R_(3) = E- Ir`
`= 8 =2 xx 1= 6V, = I.R_(3)`
Current in branch `R_(3) rArr I. = (6)/(R_(3)) = (6)/(6) = 1A`
Current in branch `R_(1) and R_(2) = I- I. = 2-1 = 1A`