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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

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`G=98Omega,(i_(g))/(i)xx100=2%`
`s=(G)/(((i)/(i_(g))-1)),:.(i)/(i_(g))=(100)/(2)=50:.S=(98)/((50-1))=2Omega`
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