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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.

Text Solution

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`R_(eq)=R_(1)+R_(2)`
= `(100pm3)Omega+(200pm4)Omega`
= `(100+200)pm(3+4)`
`300pm7Omega`
Thus the equivalent resistance is `300Omega` with a maximum permissible absolute error of `7Omega`.
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