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Question
A galvanometer has a resistance of 16Ω. It shows full scale deflection, when a current of 20 mA is passed through it. The only shunt resistance available is 0.06 which is not appropriate to convert a galvanometer into an ammeter. How much resistance should be connected in series with the coil of galvanometer, so that the range of ammeter is 8 A?
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Solution
Let ‘X’ be the resistance connected in series with galvanometer.
Since S is not sufficient for I = 10 A
G = 16 Ω,
`"I"_"g"/"I" = "S"/(("G + X") + "S")`
∴ (G + X) + S = `"I"/"I"_"g""S"`
∴ `(16 + "X" + 0.06) = 8/(2 xx 10^-2) xx 0.06`
∴ (16 + X + 0.06) = 24
∴ X = 24 - 16.06
∴ X = 7.94
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