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In a galvanometer, a current of 1µA produces a deflection of 20 divisions. It has a resistance of 10Ω. If the galvanometer has 50 divisions on its scale and a shunt of 2.5 Ω is connected across the galvanometer, the maximum current that the Galvanometer can measure now is

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$2.5 \times 10^{- 3} mA$

12.5µA

$12.5 \times 10^{- 7} A$

12.5 mA

answer is A.

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Detailed Solution

$\begin{array}{l} i_{g} = \frac{50 \times 10^{- 6}}{20} = 2 .5 \times 10^{- 6} A \\ i = (\frac{G + S}{S}) i_{g} = (\frac{12 .5}{2 .5}) \times 2 .5 \times 10^{- 6} = 12 .5 μA \end{array}$

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