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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. |
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Answer» <P> Solution :`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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