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Two resistances R(1) = 100 +- 3 Omega an...

Two resistances `R_(1) = 100 +- 3 Omega and R_(2) = 200 +- 4 Omega` are connected in series . Find the equivalent resistance of the series combination.

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`R_"eq"=R_1+R_2`
`=(100pm3)+(200pm4)Omega`
`=(100+200)pm(3+4)`
`=300pm 7 Omega`
Thus the equivalent resistance is `300 Omega` with a maximum permissible absolute error of `7 Omega`
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