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Grade 9Magnetism

In a galvanometer there is a deflection of 10 divisions per mA. The internal resistance of the galvanometer is 78 ohm. If a shunt of 2 ohm is connected to the galvanometer and there are 75 divisions in all on the scale of the galvanometer,calculate the maximum current which the galvanometer can read.

Profile image of Jitender Pal
12 Years agoGrade 9
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1 Answer

Profile image of Samiksha
7 Years ago
We will use the formula for calculating Imax of a Galvanometer, Ig = number of divisions/unit division
Ig = \frac{75}{10}, and, Imax = \frac{(Rg+Rs) Ig}{Rs} = \frac{(78+2) 75}{(78) 10} = \frac{100}{3}\boldsymbol{A}
Imax = \frac{(Rg+Rs) Ig}{Rs} = \frac{(78+2) 75}{(78) 10} = \frac{100}{3}\boldsymbol{A}

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