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A battery of 9V is connected in series w...

A battery of 9V is connected in series with resistors of 0.2`Omega,3Omega,0.4Omega` and `12Omega` respectively. How much current would flow through the 12`Omega` resistors ?

Text Solution

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`R_(s)=0.2+0.3+0.4+0.5+12Omega=13.4Omega`
`I=V/R=(9V)/(13.4Omega)=0.67A`
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