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 8V and internal resistance `1Omega` is connected as shown.

Calculate the current through each part of the circuit.

Explanation

`R _(1) =12 Omega`
`R _(1) :R_(2) : R _(3) = 1:2:3`
Let, `R _(1) = R " "R_(1) = 2 Omega`
`R _(2) = 2 R " " R _(2) = 4 Omega`
`R _(3) = 3 R " " R _(3) = 6 Omega`
According to circuit:
`((R _(1) + R _(2)) xx R _(3))/( R _(1 )+ R _(2) + R _(3))`
`R _(eq) = 3 + 1 = 4 Omega`
`I = 8/4 = 2 A`
Current through each branch is 1 Ampere.

Let current across `R _(1) is I _(1)`
Current across `R _(2)` is
` I _(2) = I _(1)`
So current across
`R _(3) = I _(3) = ( I - I _(1))`
Taking current dividing rule :
`I _(1) = ( R _(3))/( (R _(1) + R _(2) + R _(3))) xx I `
`= (6)/( (2 + 4 + 6 )) xx 1 .`
`I _(2) = I _(1) = 1 ` amp
`I _(3) = I - I _(1)`
= 1 amp