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Question
The resistance of a potentiometer wire is 8 Ω and its length is 8 m. A resistance box and a 2 V battery are connected in series with iL What should be the resistance in the box if it is desired to have a potential drop of 1 µV/mm?
Solution
Data: R = 8 Ω, L = 8 m, E = 2 V, K = 1 µV/ mm
`= 1 xx (10^-6 "V")/(10^-3 "m") = 10^-3 "V"//"m"`
K = `"V"/"L" = "ER"/(("R" + "R"_"B")"L")`, where RB is the resistance in the box.
∴ `10^-3 = (2 xx 8)/((8 + "R"_"B")8)`
∴ `8 + "R"_"B" = 2/10^-3`
∴ `8 + "R"_"B" = 2 xx 10^3`
∴ `"R"_"B" = 2000 - 8`
∴ `"R"_"B"` = 1992 ohm
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