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Q.

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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a

2.5×10−3mA

b

12.5µA

c

12.5×10−7A

d

12.5 mA

answer is A.

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Detailed Solution

ig=50×10−620=2.5×10−6Ai=G+SSig=12.52.5×2.5×10−6=12.5μA

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