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A uniform wire of resistance 12 Omegais cut into three pieces so that the ratio of the resistances R _(1) : R _(2) :R _(3) = 1 : 2 : 3and 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. |
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Answer» SOLUTION :`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 |
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