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A galvanometer has a resistance of 98Ome...

A galvanometer has a resistance of `98Omega`. If 2% of the main current is to be passed through the meter what should be the value of the shunt?

Text Solution

Verified by Experts

`G = 98 Omega , (i_g)/(I) = 2 %`
`s = (G)/(((i)/(i_g)-1)) , therefore i/(i_g) = 100/2 = 50`
`therefore S = (98)/((50-1)) = 2 Omega`
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