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A battery 9 V is connected in series wit...

A battery 9 V is connected in series with resistors of 0.2`Omega` ,0.3`Omega` 0.4 `Omega` 0.5`Omega` and 12 `Omega` respectively.How much current will flow through a 12`Omega` resistor?

Text Solution

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In series combination, same strength of current passes through each and every resistor.
`R_(eq)=R_(1)+R_(2)+R_(3)+R_(4)+R_(5)`
`=0.2+0.3+0.4+0.5+12Omega=13.4Omega`
`:.I=(V)/(R_(eq))=(9V)/(13.4Omega)=(90)/(134)`
`I=0.67` Ampere
`:.` Value of I through `12Omega` resistor is 0.67A.
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