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Two resistors of 5 Omega and 10 Omega ar...

Two resistors of `5 Omega and 10 Omega` are connected in parallel with a cell of emf 3 V and internal resistacne `1 Omega.` Calculate the current through each of the resistors.

Text Solution

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`R _(P) = ( 5 xx 10 )/( 5 + 10 ) = (50)/( 15) = (10)/(3) Omega`
`I = (E)/( R _(P) +r ) = (3)/(( 10)/(3) +1 ) = (9)/( 13) A`
`I_(1) = (IR _(2))/( R _(1) + R _(2)) = ((9)/( 13) xx 10 )/( 5 + 10 ) = (6)/(13) A`
`I _(2) = I- I _(1) = (3)/(13) A.`
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